Math 140B - Notes

Neil Donaldson

March 11, 2026

1 Continuity

The overarching goal of this course and its prequel is to make elementary calculus rigorous. We begin

with a review of some basic concepts and conventions.

Sets & Functions In these notes, essentially all functions have the form f : U → V where both U, V

are subsets of the real numbers R. To f are associated several concepts:

Domain dom( f ) = U; the inputs to f . Often implied to be the largest set on which a formula is deﬁned.

In calculus examples, the domain is typically a union of (open) intervals.

Codomain codom( f ) = V; the potential outputs of f . By convention, V = R unless necessary.

Range range( f ) = f (U) = {f (x) : x ∈ U}; the realized outputs of f and a subset of V.

Injectivity f is injective/one-to-one if f (x) = f (y) =⇒ x = y: distinct inputs produce distinct outputs.

Surjectivity f is surjective/onto if f (U) = V: all potential outputs are realized.

Inverses f is bijective/invertible if it is injective and surjective. Equivalently, ∃f

−1

: V → U satisfying

∀u ∈ U, f

−1



f (u)



= u and ∀v ∈ V, f



−1

(v)



= v

Example 1.1. The function deﬁned by f (x) =

x(x−2)

has implied

dom( f ) = R \ {0, 2} = (−∞, 0) ∪(0, 2) ∪(2, ∞)

range( f ) = (−∞, −1] ∪(0, ∞)

The function is neither injective nor surjective. By restricting the do-

main & codomain, we obtain a bijection:

dom(

f ) = [1, 2) ∪ (2, ∞)

codom(

f ) = (−∞, −1] ∪(0, ∞)

with inverse

−1

(y) =

(

1 + y

−1

y + 1 if y > 0

1 −y

−1

y + 1 if y ≤ −1

Now dom(

−1

) = codom(

f ) and codom(

−1

) = dom(

f ).

−2

−1

f (x)

−1 1 2 3

−2 −1 0 1 2

−1

(y)

Suprema and Inﬁma A set U ⊆ R is bounded above if it has an upper bound M:

∃M ∈ R such that ∀u ∈ U, u ≤ M

Axiom 1.2 (Completeness). If U ⊆ R is non-empty and bounded above, then it has a least upper

bound, the supremum of U

sup U = min



M ∈ R : ∀u ∈ U, u ≤ M



By convention, sup U := ∞ if U is unbounded above and sup ∅ := −∞; now every subset of R has a

supremum. Similarly, the inﬁmum of U is its greatest lower bound:

inf U =











max



m ∈ R : ∀u ∈ U, u ≥ m



if U = ∅ is bounded below

−∞ if U = ∅ is unbounded below

∞ if U = ∅

Examples 1.3. Here are four sets with their suprema and inﬁma stated. You should be able to verify

these assertions directly from the deﬁnitions.

U {1, 2, 3, 4} (0, 5) (−∞, π] R {

: n ∈ N}

sup U 4 5 π ∞ 1

inf U 1 0 −∞ −∞ 0

Note how the supremum/inﬁmum might or might not lie in the set itself.

Interiors, closures, boundaries and neighborhoods These last concepts might not be review, but

they will be used repeatedly.

Deﬁnition 1.4. Let U ⊆ R. A value a ∈ R is interior to U if it lies in some open subinterval of U:

∃δ > 0 such that (a −δ, a + δ) ⊆ U

A neighborhood of a is any set to which a is interior: the interval (a −δ, a + δ) is an open δ-neighborhood

of a. A punctured neighborhood of a is a neighborhood with a deleted.

The set of points interior to U is denoted U

◦

A limit point of U is the limit of some sequence (x

) ⊆ U. The closure U is the set of limit points.

The boundary of U is the set ∂U = U \U

◦

Examples 1.5. 1. If U = [1, 3), then U

◦

= (1, 3), U = [1, 3] and ∂U = {1, 3}.

2. Q

◦

= ∅ and ∂Q = Q = R.

3. (−3, 5) ∪ (5, 7] is a punctured neighborhood of 5.

1.17 Continuity of Functions

Everything in this section should be review.

Deﬁnition 1.6. A function f : U → R is continuous at u ∈ U if either/both of the following hold:

1. For all sequences (x

) ⊆ U converging to u, the sequence ( f (x

)) converges to f (u).

2. ∀ϵ > 0, ∃δ > 0 such that (∀x ∈ U),

x −u

< δ =⇒

f (x) − f (u)

< ϵ.

A function f is continuous on U if it is continuous at every point u ∈ U.

Examples 1.7. 1. We prove that f (x) = x

is continuous at u = 2.

(a) (Limit method) Let x

→ 2. By the limit laws (i.e. lim(x

) =

(

lim x

)

lim f (x

) = lim x



lim x



= 2

= f (2)

(b) (ϵ–δ method) Let ϵ > 0 be given and let δ = min





x −2

< δ =⇒

x −2

< 1 =⇒ 1 < x < 3

from which



−2



x −2



+ 2x + 2



< 19

x −2

≤ ϵ

where we used the triangle inequality.

2. Let g(x) =

(

x sin

if x = 0,

0 if x = 0

Then g is continuous at x = 0. Again this can be done with limits

or an ϵ–δ argument; both are essentially the squeeze theorem.

3. The function deﬁned by

h(x) =

(

1 + 2x

if x < 1

2 − x if x ≥ 1

is discontinuous at x = 1.

(a) The sequence with x

= 1 −

converges to 1, yet

lim h(x

) = 3 = 1 = h(1)

(b) Choose ϵ = 1 and suppose δ > 0 is given. Now choose

x = max{1 −

√

} to see that

x −1

< δ and

h(x) −h(1)

≥ 1 = ϵ

g(x)

h(x)

0 1 2

Theorem 1.8. The two parts of Deﬁnition 1.6 are equivalent.

Proof. (1 ⇒ 2) We prove the contrapositive. Suppose condition 2 is false; that is,

∃ϵ > 0, such that ∀δ > 0, ∃x ∈ U with

x −u

< δ and

f (x) − f (u)

≥ ϵ

In particular, for any n ∈ N we may let δ =

to obtain

∃ϵ > 0, such that ∀n ∈ N, ∃x

∈ U with

−u

and

f (x

) − f (u)

≥ ϵ

The sequence (x

) shows that condition 1 is false:

• ∀n,

−u

whence x

→ u.

• ∀n,

f (x

) − f (u)

≥ ϵ > 0, whence f (x

) does not converge to f (u).

(2 ⇒ 1) Suppose condition 2 is true, that (x

) ⊆ U converges to u and that ϵ > 0 is given. Then

∃δ > 0 such that

x −u

< δ =⇒

f (x) − f (u)

< ϵ

However, by the deﬁnition of convergence (x

→ u),

∃N ∈ N such that n > N =⇒

−u

< δ =⇒

f (x

) − f (u)

< ϵ

Otherwise said, f (x

) → f (u).

Rather than use these deﬁnitions every time, it is helpful to have a working dictionary.

Theorem 1.9 (Dictionary of Common Continuous Functions).

1. Suppose f and g are continuous at u and that k is constant. Then the following are continuous

at u (if deﬁned—don’t divide by zero!):

f + g, f − g, f g,

, k f , max( f , g), min( f , g)

2. If f is continuous at u and h is continuous at f (u) , then h ◦ f is continuous at u.

3. Algebraic functions are continuous: these are functions constructed using ﬁnitely many addi-

tion/subtraction, multiplication/division and n

root operations.

4. The familiar transcendental functions are continuous: exp, ln, sin, etc.

Example 1.10. f (x) = sin

√

x−2

+ cos

−1

is continuous on its domain (−∞, 0) ∪ (0, 1) ∪ (1, ∞).

Theses claims are tedious to prove using elementary deﬁnitions. In particular, it is better to defer a

proof of the transcendental claim until we can deﬁne such functions using power series, after which

continuity comes for free.

Exercises 1.17. Key concepts/results: Suprema/Completeness, Sequential & ϵ-δ continuity

1. Give examples to show that g ◦ f being continuous can happen with:

(a) f continuous and g discontinuous. (b) g continuous and f discontinuous.

You may use pictures, but make sure they clearly describe the functions f , g.

2. (a) Prove that the function f (x) = x

is continuous at x = −2 using an ϵ–δ argument.

(b) Prove that f (x) = x

is continuous at x = u using an ϵ–δ argument.

3. Prove that the following are discontinuous at x = 0: use both deﬁnitions of continuity.

(a) f (x) = 1 for x < 0 and f (x) = 0 for x ≥ 0.

(b) g(x) = sin(1/x) for x = 0 and g(0) = 0.

4. If f is continuous at u, prove that it is bounded on some set (u −δ, u + δ) ∩dom( f ).

5. Prove the following parts of Theorem 1.9 using ϵ–δ arguments.

(a) If f , g are continuous at u, then f − g is continuous at u.

(b) If f , g are continuous at u, then f g is continuous at u.

6. Suppose f : U → R is a function whose domain U contains an isolated point a: i.e. ∃r > 0 such

that (a −r, a + r) ∩U = {a}. Prove that f is continuous at a.

7. Refresh your prerequisites by giving formal proofs:

(a) (Suprema and sequences) If M = sup U, then ∃(x

) ⊆ U such that x

→ M.

(This has to work even if M = ∞!)

(b) (Limit of a bounded sequence) If (x

) ⊆ [a, b] and x

→ x, then x ∈ [a, b].

(Hint: If (x

) ⊆ [a, b], explain why there exist intervals I

⊇ I

⊇ ··· such that inﬁnitely

many (x

) lie in each interval I

. Hence obtain a subsequence (x

) and prove that it is Cauchy.

)

8. (Very Hard) Consider the function f : R → R where

f (x) =

(

whenever x =

∈ Q with q > 0 and gcd(p, q) = 1

0 if x ∈ Q

For example, f (1) = f (2) = f (−7) = 1, and f (

) = f (−

) = f (

) = ··· =

, etc. Prove that f

is continuous at each point of R \Q and discontinuous at each point of Q.

(Hint: for continuity, consider A = {r ∈ Q : f (r) ≥

} where q ≥

. . . )

This is a good moment to review Cauchy completeness: that a sequence is convergent if and only if it is Cauchy:

∀ϵ > 0, ∃N such that m, n > N =⇒

− x

< ϵ

1.18 Properties of Continuous Functions

In this section we describe the behavior of a continuous function on an interval. We ﬁrst consider the

special case when the domain is a closed bounded interval [a, b] .

Theorem 1.11 (Extreme Value Theorem). A continuous function on a closed, bounded interval is

bounded and attains its bounds. Otherwise said, if f : [a, b] → R is continuous, then

∃x, y ∈ [a, b] such that f (x) = sup range( f ) and f (y) = inf range( f )

In particular, the supremum and inﬁmum are ﬁnite.

Proof. Suppose f is continuous with domain [a, b] and let M = sup{f (x) : x ∈ [a, b]}. We invoke the

three parts of Exercise 1.17.7:

• (Part a) There exists a sequence (x

) ⊆ [a, b] such that f (x

) → M.

• (Part c) There exists a convergent subsequence (x

) with limit x.

• (Part b) x ∈ [a, b].

Since f is continuous, we now have f (x) = lim

k→∞

f (x

) = M. This shows that M is ﬁnite and that f

attains its least upper bound. For the lower bound, apply this to −f .

It is worth considering how the result can fail when one of the hypotheses is weakened. For example:

f discontinuous f : [0, 1] → R : x 7→

(

x if x = 1

0 if x = 1

is bounded but does not attain its bounds.

dom( f ) not closed f : [0, 1) → R : x 7→ x is bounded but does not attain its bounds.

dom( f ) not bounded f : [0, ∞) → R : x 7→ x is unbounded.

We now consider continuous functions on arbitrary intervals. The next result should be familiar from

elementary calculus and is intuitively obvious from the na

ıve notion of continuity (draw the graph

without taking your pen off the page).

Theorem 1.12 (Intermediate Value Theorem). Let f : I → R be continuous on an interval I. Sup-

pose a, b ∈ I with a < b and that f (a) = f (b). If L lies between f (a) and f (b), then ∃ξ ∈ (a, b) such

that f (ξ) = L.

Example 1.13. Let f (x) = cos x with a =

, b = 3π

and L =

; then

f (ξ) = L ⇐⇒ ξ ∈



5π

7π



There may therefore be several suitable values of ξ. It is

even possible (Exercise 2) for there to be inﬁnitely many.

−1

f (x)

a bπ 2π

Proof. Suppose WLOG that f (a) < L < f (b) and let

S = {x ∈ [a, b] : f (x) < L}

Plainly S ⊆ [a, b) is non-empty, hence ξ := sup S exists

and ξ ∈ [a, b]. It remains to show that ξ satisﬁes the

required properties.

By Exercise 7, ∃(s

) ⊆ S with lim s

= ξ. Since f is

continuous, f (ξ) = lim f (s

) ≤ L. In particular, ξ = b.

a b

f (a)

f (b)

To ﬁnish, play a similar game with the sequence deﬁned by t

= min{b, ξ +

} (see Exercise 4).

Example 1.14. The intermediate value theorem is useful for demonstrating the existence of solutions

to equations. For example, we show that the equation x2

= 1 has a solution.

• Observe that g(x) = x2

−1 is continuous.

• g(0) = −1 < 0.

• g(1) = 1 > 0.

• By the intermediate value theorem ∃ξ ∈ (0, 1) such

that g( ξ) = 0: that is ξ · 2

= 1.

−1

g(x)

It is inefﬁcient, but one can home in on ξ by repeatedly halving the size of the interval: for instance,

) =

√

−1 < 0, g(

) =

·2

3/4

−1 ≈ 0.26 > 0 . . . =⇒

< ξ <

Corollary 1.15. Continuous functions map intervals to intervals (or points).

Proof. An interval I is characterized by the following property

∀x

, x

∈ I, x ∈ R, x

< x < x

=⇒ x ∈ I

Let f : I → R be continuous and suppose its range f (I) is not a single point. If f (a) < L < f (b), then

∃ξ between a, b such that f (ξ) = L. Otherwise said, L ∈ f (I) and so f (I) is an interval.

More generally, if dom( f ) =

is written as a union of disjoint intervals and f is continuous, then

range( f ) =

[

f (I

)

is also a union of intervals, though these need not be disjoint: a continuous function can bring inter-

vals together, but cannot break them apart.

Example 1.16. The function f (x) =

√

−4 has implied domain (−∞, −2] ∪[2, ∞) and range [0, ∞).

Both halves of the domain are mapped onto the same interval range( f ) .

Exercises 1.18. Key concepts: Extreme Value Theorem, Intermediate Value Theorem

Continuous functions preserve intervals

1. Give examples of the following:

(a) An unbounded discontinuous function on a closed bounded interval.

(b) An unbounded continuous function on a non-closed bounded interval.

bounds.

2. Consider the function f (x) =

(

x sin

if x = 0

0 if x = 0

(a) Explain why f is continuous on any interval I.

(b) Suppose a < 0 < b and that f (a), f (b) have opposite signs. If L = 0, show that the

intermediate value theorem is satisﬁed by inﬁnitely many distinct values ξ.

3. Use the intermediate value theorem to prove that the equation 8x

−12x

−2x + 1 = 0 has at

least 3 real solutions (and thus, by the fundamental theorem of algebra, exactly 3).

4. Complete the proof of the intermediate value theorem by deﬁning t

= min(b, ξ +

5. (a) Suppose f : U → R is continuous and that U =

k=1

is the union of a ﬁnite sequence (I

)

of closed bounded intervals. Prove that f is bounded and attains its bounds.

(b) Let U =

∞

n=1

, where I

= [

2n−1

] for each n ∈ N. Give an example of a continuous

function f : U → R which is either unbounded or does not attain its bounds. Explain.

More generally, if f : U → V is a continuous function between topological spaces and a, b lie in the same component of

U, then f (a), f (b) lie in the same component of f (U). In our examples each component is an interval.

1.19 Uniform Continuity

Recall Deﬁnition 1.6: f : U → R is continuous at all points

y ∈ U provided

∀y ∈ U, ∀ϵ > 0, ∃δ > 0 such that (∀x ∈ U)

x −y

< δ =⇒

f (x) − f (y)

< ϵ

Note the order of the quantiﬁers: δ is permitted to depend both on y and ϵ. In the na

ıve sense of

continuity (x close to y =⇒ f (x) close to f (y)), the meaning of close can depend on the location y.

Uniform continuity is a stronger condition where the meaning of close is independent of location.

Deﬁnition 1.17. f : U → R is uniformly continuous if

∀ϵ > 0, ∃δ > 0 such that (∀x, y ∈ U)

x −y

< δ =⇒

f (x) − f (y)

< ϵ

We’ve included the (typically) hidden quantiﬁers (∀x, y) to make clear that δ is independent of x, y.

Note also that the deﬁnition is now symmetric in x, y.

Example 1.18. Consider f (x) =

1. If 0 < a < b ≤ ∞, then f is uniformly continuous on [a, b).

Let ϵ > 0 be given and let δ = a

ϵ. Then ∀x, y ∈ [a, b),

x −y

< δ =⇒



−



y − x



≤

= ϵ

2. If 0 < b ≤ ∞, then f is not uniformly continuous on (0, b).

Let ϵ = 1 and suppose δ > 0 is given.

Let x = min(δ, 1,

) and y =

Certainly x, y ∈ (0, b) and

x −y

≤

< δ. However,

f (x) − f (y)

≥ 1 = ϵ

f (x)

a b

Think about how ϵ and δ must relate as one slides the intervals in the picture up/down and left/right.

Some intuition will help make sense of the example.

Bounded/unbounded gradient In part 1 ϵ = δa

, where

′

(a)

bounds the gradient of f .

By contrast, the slope of f is unbounded in part 2.

Extensibility In part 1 the domain of f may be extended to a (and to b if ﬁnite): g : [a, b] → R : x 7→

is continuous. In part 2, this is impossible: there is no continuous function g : [0, b) → R such

that g(x) =

whenever x > 0.

Informally, if a continuous function f has bounded gradient, or if you can ‘ﬁll in the holes’ at the

endpoints of dom( f ), then f is uniformly continuous. When uniform continuity is used abstractly in

a proof, it is often one of the above properties that is being invoked. The remainder of this section

involves making these observations watertight.

To promote symmetry, we use y instead of u for a generic point of dom( f ).

Theorem 1.19. Let f : I → R be continuous on an interval I and differentiable with bounded

derivative on the interior I

◦

. Then f is uniformly continuous on I.

The proof depends on the mean value theorem, which should be familiar from elementary calculus;

we’ll discuss a proof later on.

Proof. Suppose

′

(x)

≤ M on I

◦

. Let ϵ > 0 be given, let δ =

and suppose (y, x) ⊆ I. Then

x −y

< δ =⇒ ∃ξ ∈ I

◦

such that f

′

(ξ) =

f (x) − f (y)

x −y

(MVT)

=⇒

f (x) − f (y)



′

(ξ)



x −y

< Mδ = ϵ

Theorem 1.19 isn’t a biconditional: for instance, Exercise 1.19.5 shows that f (x) =

√

x on [0, ∞) and

g(x) = x

1/3

on R are both uniformly continuous even though both have unbounded slope.

We now discuss extensibility and how uniform continuity relates to continuity on closed sets. First

we see that for closed bounded sets, uniform continuity is nothing new.

Theorem 1.20. If g : [a, b] → R is continuous, then it is uniformly continuous.

Proof. Suppose g is continuous but not uniformly so. Then

∃ϵ > 0 such that ∀δ > 0, ∃x, y ∈ [a, b] for which

x −y

< δ and

g(x) − g(y)

≥ ϵ (∗)

For each n ∈ N, let δ =

to see that there exists sequences (x

), (y

) ⊆ [a, b] satisfying the above.

By Bolzano–Weierstraß, the bounded sequence (x

) has a convergent subsequence x

→ x ∈ [a, b].

Clearly

−y

→ 0 =⇒ y

→ x

But then

g(x

) − g(y

)

→ 0, which contradicts (∗) .

Now we build towards a partial converse.

Lemma 1.21. If f : U → R is uniformly continuous and (x

) ⊆ U is a Cauchy sequence, then



f (x

)



is also Cauchy.

Proof. Let ϵ > 0 be given. Then:

• (Uniform Continuity) ∃δ > 0 such that

x −y

< δ =⇒

f (x) − f (y)

< ϵ.

• (Cauchy) ∃N ∈ N such that m, n > N =⇒

− x

< δ.

Putting these together, we see that

∃N ∈ N such that m, n > N =⇒

f (x

) − f (x

)

< ϵ

Otherwise said,



f (x

)



is Cauchy.

We now see that a function f : I → R is uniformly continuous on a bounded interval if and only if it

has a continuous extension g : I → R deﬁned on the closure of its domain.

Theorem 1.22. Suppose f : I → R is continuous where I is a bounded interval with endpoints a < b.

Deﬁne g : [a, b] → R via

g(x) =











f (x) if x ∈ I

lim f (x

) whenever (x

) ⊆ I and x

→ a

lim f (x

) whenever (x

) ⊆ I and x

→ b

Then f is uniformly continuous if and only g is well-deﬁned (g is continuous, if well-deﬁned).

Proof. (⇒) Suppose f is uniformly continuous on I and that a ∈ I. Let (x

), (y

) ⊆ I be sequences

converging to a. To show that g is well-deﬁned, we must prove that



f (x

)



and



f (y

)



are

convergent, and to the same limit. For this, we deﬁne a sequence

) = (x

, y

, x

, y

, x

, y

, . . .)

Since (x

) and (y

) have the same limit a, we conclude that u

→ a. But then (u

) is Cauchy.

By Lemma 1.21,



f (u

)



is also Cauchy and thus convergent. Since



f (x

)



and



f (y

)



are

subsequences of a convergent sequence, they must also converge to the same (ﬁnite) limit.

The argument when b ∈ I is identical.

(⇐) If g is well-deﬁned then it is continuous (Deﬁnition 1.6, part 1); by Theorem 1.20 it is uniformly

so. Since f = g on a subset of dom(g), the same choice of δ will work for f as for g: f is therefore

uniformly continuous.

Examples 1.23. 1. Consider f : x 7→ x

(a) If dom( f ) is the open interval (−3, 10), then f is uniformly continuous since its derivative

′

(x) = 2x is bounded (

′

(x)

≤ 20). The continuous extension is g(x) = x

on [−3, 10].

(b) If dom( f ) is the inﬁnite interval (−3, ∞), then neither Theorem 1.19 nor 1.22 applies: both

′

and the domain (−3, ∞) are unbounded.

Instead, note that if ϵ = 1, then for any δ > 0, we can choose x =

and y =

. Clearly

x −y

< δ and



−y



= 1 +

> 1 = ϵ

whence f is not uniformly continuous.

2. f (x) = x sin

is continuous on the interval ( 0, ∞). Strictly, neither Theorem 1.19 nor 1.22 apply

since the derivative

′

(x) = sin

−

cos

is unbounded as is the domain. However, by breaking the domain into two pieces. . .

• On [1, ∞), the derivative is bounded:

′

(x)

≤ 1 +

≤ 2 by the triangle inequality.

Theorem 1.19 says f is uniformly continuous on [1, ∞).

• f is continuous on (0, 1] and, by the squeeze theorem

→ 0

=⇒ lim f (x

) = 0

Extending f so that f (0) = 0 deﬁnes a continuous extension. By Theorem 1.22, f is uni-

formly continuous on ( 0, 1].

Putting this together (Exercise 6), f is uniformly continuous on (0, ∞). Indeed the function

h(x) =

(

x sin

if x = 0

0 if x = 0

is uniformly continuous on R.

Exercises 1.19. Key concepts: Uniform Continuity (same δ for all locations), Bounded gradient,

Continuous extensions

1. Which functions are uniformly continuous? Justify your answers.

(a) f (x) = x

on [−1, 1] (b) f (x) = x

on (−1, 1]

−4

on ( 0, 2] (d) f (x) = x

−4

on ( 1, 2]

(e) f (x) = x

sin

on ( 0, 1]

2. Prove that each function is uniformly continuous by verifying the ϵ–δ property.

(a) f (x) = 2x −14 on R (b) f (x) = x

on [1, 5]

−1

on ( 1, ∞) (d) f (x) =

x+1

x+2

on [0, 1]

3. Prove that f (x) = x

is not uniformly continuous on R.

4. (a) Suppose f is uniformly continuous on a bounded interval I. Prove that f is bounded on I.

(b) Use part (a) to write down a bounded interval on which the function f (x) = tan x is

deﬁned, but not uniformly continuous.

5. Both parts of this question are easy using Exercise 6. Do them explicitly using the ϵ–δ property.

(a) Let f (x) =

√

x with domain [0, ∞). Show that f

′

(x) is unbounded, but that f is still

uniformly continuous on [0, ∞).

(Hint: let δ = (

)

and consider the cases x ≥ y ≥ 0, x ≤ y ≤ 0 and x > 0 > y separately)

(b) Prove that g(x) = x

1/3

is uniformly continuous on R.

(Hint: let δ = ϵ

and WLOG assume 0 ≤ y ≤ x. Now compute (

√

y + ϵ)

. . . )

6. Suppose f is uniformly continuous on intervals U

, U

for which U

∩U

is non-empty. Prove

that f is uniformly continuous on U

∪U

(Hint: if x, y do not lie in the same U

, U

, choose some a ∈ U

∩U

between x and y)

1.20 Limits of Functions

In elementary calculus you likely saw many calculations of the following form:

lim

x→3

−9

x −3

= lim

x→3

(x −3)(x + 3)

x −3

= lim

x→3

(x + 3) = 6

Loosely speaking, this means that if (x

) ⊆ R \ {3} is a sequence converging to 3, then



f (x

)



converges to 6. Our goal in this section is to make this notation precise.

Deﬁnition 1.24. Suppose f : U → R, that S ⊆ U, and that a is the limit of some sequence in S.

We say that L is the limit of f (x) as x tends to a along S, written lim

x→a

f (x) = L, provided

∀(x

) ⊆ S, lim x

= a =⇒ lim f (x

) = L

We can now deﬁne one-sided and two-sided limits of functions:

Right-hand limit: lim

x→a

f (x) = L means ∃S = (a, b) ⊆ U for which lim

x→a

f (x) = L

Left-hand limit: lim

x→a

−

f (x) = L means ∃S = (c, a) ⊆ U for which lim

x→a

f (x) = L

Two-sided limit: lim

x→a

f (x) = L means ∃S = (c, a) ∪ (a, b) ⊆ U for which lim

x→a

f (x) = L

• If U = dom( f ) is unbounded, then the one-sided deﬁnitions apply when a = ±∞. We omit the

± modiﬁers: for instance,

lim

x→∞

f (x) = L ⇐⇒ lim

x→∞

f (x) = L for some S = (c, ∞) ⊆ U

• Note that f need not be deﬁned at a, though U = dom( f ) must contain at least some punctured

neighborhood of a (one-sided for a one-sided limit). This will certainly happen if U is a union

of intervals of positive length. In such a case, one may simply replace S with U \ {a} in the

deﬁnition: this is precisely what we did in the motivating example where U = R \ {3}.

Moreover, in such a situation, Deﬁnition 1.6 recovers a familiar idea from elementary calculus:

f is continuous at a ∈ U ⇐⇒ f (a) =







lim

x→a

f (x) when a ∈ U

◦

lim

x→a

f (x) when a ∈ U \U

◦

(∗)

Warning! When dom( f ) does not contain a punctured neighborhood of a, the right hand side

doesn’t exist and the assertion is false!

• By modifying the proof of Theorem 1.8 when a, L ∈ R are ﬁnite, we can restate using ϵ-

language. For instance, lim

x→a

f (x) = L means

∀ϵ > 0, ∃δ > 0 such that (∀x ∈ dom f ) 0 <

x −a

< δ =⇒

f (x) − L

< ϵ

If a and/or L is inﬁnite, use the language of unboundedness: e.g., lim

x→a

f (x) = ∞ means

∀M > 0, ∃δ > 0 such that 0 <

x −a

< δ =⇒ f (x) > M

There are ﬁfteen distinct combinations: three two-sided and six each of the one-sided limits!

Examples 1.25. 1. Let f (x) =

2+x

where dom( f ) = U = R \ {0} = (−∞, 0) ∪ (0, ∞)

The following should be clear:

lim

x→3

f (x) =

lim

x→∞

f (x) = 1

To compute the ﬁrst, for instance, we could choose S = (0, 3) ∪(3, ∞); if (x

) ⊆ S and x

→ 3,

then the limit laws justify the ﬁrst claim

lim

n→∞

f (x

) =

2 + 3

as does the fact that f is continuous at x = 3. The second claim can be checked similarly.

We can take one-sided limits at x = 0:

lim

x→0

f (x) = ∞ and lim

x→0

−

f (x) = −∞

For instance, let (x

) ⊆ (0, ∞) satisfy x

→ 0. Again, the

limit laws show that lim

n→∞

f (x

) = ∞, which is enough to

justify the ﬁrst claim.

Finally, the sequences deﬁned by x

and y

= −

both lie in S = R \ {0} and converge to zero, yet

lim

n→∞

f (x

) = ∞ = −∞ = lim

n→∞

f (y

)

It follows that the two-sided limit lim

x→0

f (x) does not exist.

−9

−6

−3

f (x)

−2 −1 1 2

f (x

)

f (y

)

2. Let f (x) =

whenever x = 0 and additionally let f (0) = 0. Here the two-sided limit exists

lim

x→0

f (x) = ∞

However the value of the function at x = 0 does not equal this limit: clearly f is discontinuous

at x = 0.

3. We revisit our motivating example. Let f (x) =

−9

x−3

have domain U = R \ {3}. Whenever

= 3, we see that

f (x

) =

−3)(x

+ 3)

−3

= x

+ 3

By the limit laws, we conclude that lim f (x

) = 3 + 3 = 6 and so

lim

x→3

−9

x −3

= 6

Since we referenced the limit laws for sequences so often in the examples, it is appropriate to update

them to this new context. We do so without proof.

Corollary 1.26 (Limit Laws for functions). Suppose f , g : U → R satisfy L = lim

x→a

f (x) and M =

lim

x→a

g(x) exist. Then,

1. lim

x→a

( f + g) (x) = L + M.

2. lim

x→a

( f g)(x) = LM.

3. lim

x→a





(x) =

(requires M = 0).

4. If L ∈ R and h is continuous at L, then lim

x→a

(h ◦ f )(x) = h(L).

5. (Squeeze Theorem) If L = M and f (x) ≤ h(x) ≤ g(x) for all x ∈ U, then lim

x→a

h(x) = L.

The corresponding results for one-sided limits also hold.

As with the original limit laws for sequences, parts 1–3 apply provided the limits are not indeterminate

forms (e.g. ∞ −∞, 0 · ∞,

∞

). We’ll see later how l’H

opital’s rule may be applied to such cases.

Examples 1.27. 1. Since f (x) =

−2

is a rational function (continuous at all points of its domain),

we quickly conclude that

lim

x→2

+ 5

−2

= f (2) =

Alternatively, we may tediously invoke the other parts of the theorem:

lim

x→2

+ 5

−2

(3)

lim(x

+ 5)

lim( 3x

−2)

(1)

lim x

+ lim 5

lim 3x

−lim 2

(2)

(lim x)

+ 5

(lim 3)(lim x)

−2

+ 5

3 ·2

−2

2. As x → ∞, the simplistic approach results in a nonsense indeterminate form:

lim

x→∞

+ 5

−2

lim(x

+ 5)

lim( 3x

−2)

∞

However, a little pre-theorem algebra quickly yields

lim

x→∞

+ 5

−2

= lim

x→∞

1 + 5x

−2

3 −2x

−2

lim( 1 + 5x

−2

)

lim( 3 −2x

−2

)

Be careful! The expressions

−2

and

1+5x

−2

3−2x

−2

do not describe the same function, yet their limits at ∞ are equal. The

ease of equating these limits is one of the advantages of the ‘∃S’ formulation in Deﬁnition 1.24. Think about why; what is

a suitable set S in this context?

Classiﬁcation of Discontinuities

We now consider the ways in which a function can fail to be continuous.

Deﬁnition 1.28. Suppose that a function is continuous on an interval except at ﬁnitely many values:

we call these isolated discontinuities.

Examples 1.29. 1. f (x) =

has a discontinuity at x = 0 since it is continuous on the interval R,

except at one point x = 0. Note that a function need not be deﬁned at a discontinuity!

2. f (x) =

sin

has a non-isolated discontinuity at x = 0: on any interval containing zero, f has

inﬁnitely many discontinuities: x =

πn

where

∈ N.

The next result helps us classify isolated discontinuities.

Theorem 1.30. Let f be deﬁned on a punctured neighborhood of a ∈ R. Then

lim

x→a

f (x) = L ⇐⇒ lim

x→a

f (x) = L = lim

x→a

−

f (x)

Proof. (⇒) Let S = (c, a) ∪ (a, b) satisfy the deﬁnition for lim

x→a

f (x) = L. Since any sequence (say) in

is also in S, plainly S

= (a, b) and S

−

= (c, a) satisfy the one-sided deﬁnitions.

(⇐) Suppose S

−

= (c, a) and S

= (a, b) satisfy the one-sided deﬁnitions and denote S = S

−

∪ S

Let (x

) ⊆ S be such that x

→ a. Clearly (x

) is the disjoint union of two subsequences

) ∩ S

and (x

) ∩ S

−

, both of which

converge to a. There are three cases:

L ﬁnite: Let ϵ > 0 be given. Because of the one-sided limits,

• ∃N

such that n > N

and x

> a =⇒

f (x

) − L

< ϵ

• ∃N

such that n > N

and x

< a =⇒

f (x

) − L

< ϵ

Now let N = max(N

, N

) in the deﬁnition of limit to see that lim f (x

) = L. Since this

holds for all sequences (x

) ⊆ S converging to a, we conclude that lim

x→a

f (x) = L.

L = ±∞: This is an exercise.

Example 1.31. Recalling elementary calculus, we show that the following is continuous at x = 1:

f (x) =

(

−3 if x ≥ 1

3 −5x if x < 1

Step 1: Compute the left- and right-handed limits and check that these are equal:

lim

x→1

−

f (x) = lim

x→1

−

3 −5x = −2, lim

x→1

f (x) = lim

x→1

−3 = −2

Step 2: Check that the value of the limits equals that of the function: f (1) = 1

−3 = −2.

It is possible for one of these subsequences to be ﬁnite; say if x

> a for all large n. This is of no concern; one of the ϵ-N

conditions would be empty and thus vacuously true.

Recalling (∗) on page 13, we describe the different types of isolated discontinuity at some point a.

Removable discontinuity The two-sided limit lim

x→a

f (x) = L is ﬁ-

nite, and either:

f (a) = L or f (a) is undeﬁned.

The term comes from the fact that we can remove the discon-

tinuity by changing the behavior of f only at x = a:

f (x) :=

(

f (x) if x = a

lim

x→a

f (x) if x = a

is now continuous at x = a. In the pictures,

(x) =

−9

x −3

and f

(x) =

(

x sin(

) if x = 0

1 if x = 0

have removable discontinuities at x = 3 and 0 respectively.

(x)

Jump Discontinuity The one-sided limits are ﬁnite but not equal. A

jump discontinuity cannot be removed by changing or insert-

ing a value at x = a. The picture shows

g(x) =

(

1 if x > 0

−1 if x < 0

with a jump discontinuity at x = 0.

g(x)

Inﬁnite discontinuity The one-sided limits exist but at least one is

inﬁnite. We call the line x = a a vertical asymptote. The picture

shows

h(x) =

with an inﬁnite discontinuity x = 0. The fact that the one-

sided limits of h are equal (and inﬁnite) is irrelevant.

h(x)

Essential discontinuity At least one of the one-sided limits does

not exist. The picture shows j(x) = sin

for which neither of

the limits lim

x→0

j(x) exist.

j(x)

It is also reasonable to refer to removable, inﬁnite or essential discontinuities at interval endpoints.

Exercises 1.20. Key concepts: lim

x→a

f (x) = L, ϵ, δ, M, N versions, Limit Laws, Discontinuities

1. Given f (x) =

, ﬁnd lim

x→∞

f (x), lim

x→−∞

f (x), lim

x→0

−

f (x), lim

x→0

f (x) and lim

x→0

f (x), if they exist.

2. Evaluate the following limits using the methods of this section

(a) lim

x→a

√

x −

√

x −a

(b) lim

x→a

−3/2

− a

−3/2

x −a

x→0

√

1 + 3x

−1

(d) lim

x→−∞

√

4 + 3x

−2

3. Suppose that the limits L = lim

x→a

f (x) and M = lim

x→a

g(x) exist.

(a) Suppose f (x) ≤ g(x) for all x in some interval (a, b). Prove that L ≤ M.

(b) Do we have the same conclusion if we have f (x) < g(x) on (a, b), or can we conclude that

L < M? Prove your assertion, or give a counter-example.

4. Suppose that lim

x→∞

f (x) = lim

x→∞

g(x) = ∞. Using only this information, which of the following

can you always evaluate? Prove your assertions in each case.

(a) lim

x→∞

( f + g) (x) (b) lim

x→∞

( f − g) (x) (c) lim

x→∞

( f g)(x) (d) lim

x→∞

( f /g)(x)

5. Complete the proof of Theorem 1.30 by considering the L = ±∞ cases.

6. Graph f : R → R, ﬁnd and identify the types of its discontinuities.

f (x) =











0 x = 0, ±1

0 <

< 1

> 1

7. Find the discontinuities and identify their types for the following function

f (x) =

(

sin

if x < 0 or x > 1

if 0 < x ≤ 1

8. Verify the claim following Deﬁnition 1.24: lim

x→a

f (x) = L if and only if

∀ϵ > 0, ∃δ > 0 such that 0 <

x −a

< δ =⇒

f (x) − L

< ϵ

9. Recall Exercise 1.17.6, where we saw that a function f : U → R is continuous at any isolated

point a ∈ U.

(a) Any function with domain dom( f ) = Z is continuous everywhere! Explain why we

cannot deﬁne any limits lim

x→a

(±)

f (x) for such a function.

(Hint: Being unable to deﬁne a limit is different from saying lim f (x) = DNE: see page 13.)

(b) Suppose g(x) = x

h(x) has dom(g) = {0} ∪ {

: n ∈ Z}, where h is any function taking

values in the interval [−1, 1]. Explain why g is continuous at every point of its domain.

(These awkward examples of continuity can be avoided if we follow our usual approach where a domain

is a union of intervals of positive length. This restriction is essentially baked in to the Deﬁnition 1.24.)

2 Sequences and Series of Functions

If ( f

) is a sequence of functions, what should we mean by lim f

? This question is of huge relevance

to the history of calculus: Issac Newton’s work in the late 1600’s made great use of power series, which

are naturally constructed as limits of sequences of polynomials.

For instance, for each n ∈ N

, we might consider the polynomial function f

: R → R deﬁned by

(x) =

∑

k=0

= 1 + x + ···+ x

This is easily differentiated and integrated using the power law. What, however, are we to make of

the series

f (x) :=

∞

∑

n=0

= 1 + x + x

+ ··· ?

Does this make sense as a function? What is its domain? Does it equal the limit of the sequence

( f

) in any meaningful way? Is it continuous, differentiable, integrable? If so, can we compute its

derivative or integral term-by-term: for instance, is it legitimate to write

′

(x) =

∞

∑

n=1

n−1

= 1 + 2x + 3x

+ ··· ?

To many in Newton’s time, such technical questions were less important than the application of cal-

culus to the natural sciences. For the 18

and 19

century mathematicians who followed, however,

the widespread application of calculus only increased the imperative to rigorously address these

issues.

2.23 Power Series

First we review some of the important deﬁnitions, examples and results concerning inﬁnite series.

Deﬁnition 2.1. Let (b

)

∞

n=m

be a sequence of real numbers. The (inﬁnite) series

∑

is the limit of the

sequence (s

) of partial sums,

∑

k=m

= b

+ b

m+1

+ ···+ b

∞

∑

n=m

= lim

n→∞

The series

∑

is said to converge, diverge to inﬁnity or diverge by oscillation

as does (s

∑

is absolutely convergent if

∑

converges. A convergent series that is not absolutely convergent

is conditionally convergent.

Recall that every sequence (s

) has subsequences tending to each of

lim sup s

= lim

N→∞

sup{x

: n > N} and lim inf s

= lim

N→∞

inf{x

: n > N}

If (s

) converges, or diverges to ±∞, then lim s

= lim sup s

= lim inf s

. The remaining case, divergence by oscillation,

is when lim inf s

= lim sup s

: there exist (at least) two subsequences tending to different limits.

Examples 2.2. These examples form the standard reference dictionary for analysis of more compli-

cated series. Make sure they are familiar!

1. (Geometric series) If r is constant, then s

∑

k=0

1−r

n+1

1−r

. It follows that

∞

∑

n=0











converges (absolutely) to

1−r

if − 1 < r < 1

diverges to ∞ if r ≥ 1

diverges by oscillation if r ≤ −1

2. (Telescoping series) If b

n(n+1)

, then s

∑

k=1

= 1 −

n+1

=⇒

∞

∑

n=1

n(n+1)

= 1.

∞

∑

n=1

is (absolutely) convergent. In fact

∞

∑

n=1

, though checking this explicitly is tricky.

4. (Harmonic series)

∞

∑

n=1

is divergent to ∞.

5. (Alternating harmonic series)

∞

∑

n=1

(−1)

is conditionally convergent.

Theorem 2.3 (Root Test). Given a series

∑

, let β = lim sup

1/n

• If β < 1 then the series converges absolutely.

• If β > 1 then the series diverges.

We give sketch proofs or refer to a standard ‘test.’ Review these if you are unfamiliar.

1. s

−rs

= 1 + r + ··· + r

−(r + ··· + r

+ r

n+1

) = 1 −r

n+1

=⇒ s

1−r

n+1

1−r

2. By partial fractions, b

−

n+1

=⇒ s



1 −





−



+ ··· +



−

n+1



= 1 −

n+1

3. Use the comparison or integral tests. Alternatively: For each n ≥ 2, we have

n(n−1)

. By part 2,

∑

k=1

< 1 +

∑

k=1

k(k −1)

< 1 +

∞

∑

n=1

n(n −1)

= 2

Since (s

) is monotone-up and bounded above by 2, we conclude that

∑

is convergent.

4. Use the integral test. Alternatively, observe that

n+1

−s

n+1

∑

k=2

−1

≥

n+1

=⇒ s

≥

−−−→

n→∞

∞

Since s

∑

k=1

is monotone-up, we conclude that s

→ ∞.

5. Use the alternating series test, or explicitly check that both the even and odd partial sums (s

) and ( s

2n+1

) are

convergent (monotone and bounded) to the same limit (essentially the proof of the alternating series test).

Root Test: β < 1 =⇒ ∃ϵ > 0 such that

1/n

≤ 1 − ϵ (for large n) =⇒

∑

converges by comparison with

∑

(1 − ϵ)

β > 1 =⇒ some subsequence of (

1/n

) converges to β > 1 =⇒ b

↛ 0 =⇒

∑

diverges (n

-term test).

The root test is inconclusive if β = 1. Some simple inequalities

yield a test that is often easier to

apply.

Corollary 2.4 (Ratio Test). Given a series

∑

• If lim sup



n+1



< 1 then

∑

converges absolutely.

• If lim inf



n+1



> 1 then

∑

diverges.

We are now ready to properly deﬁne and analyze our main objects of interest.

Deﬁnition 2.5. A power series centered at c ∈ R with coefﬁcients a

∈ R is a formal expression

∞

∑

n=m

(x −c)

where x ∈ R is considered a variable. A power series is a function whose implied domain is the set

of x for which the resulting inﬁnite series converges.

It is common to refer simply to a series, and modify by inﬁnite/power only when clarity requires.

Almost always m = 0 or 1, and it is common for examples to be centered at c = 0.

Example 2.6. By the geometric series formula,

∞

∑

n=0

(−1)

(x −4)

1 −

−(x−4)

x −2

whenever



−

x −4



< 1 ⇐⇒ 2 < x < 6

The series is valid (converges) only on the subinterval (2, 6) of

the implied domain of the function x 7→

x−2

The behavior as x → 2

is unsurprising, since evaluating the

power series results in the divergent inﬁnite series

∑

1 = +∞

By contrast, as x → 6

−

we see that limits and inﬁnite series do

not interact as we might expect,

lim

x→6

−

∞

∑

n=0

(−1)

(x −4)

= lim

x→6

−

x −2

∞

∑

n=0

lim

x→6

−

(−1)

(x −4)

∑

(−1)

= DNE

with the last series being divergent by oscillation.

−6

−4

−2

−2 2 4 6

Power Series

f (x) =

x−2

The example shows that we cannot blindly take limits inside an inﬁnite sum; understanding precisely

when this is possible is one of our primary goals.

You should have encountered these previously: lim inf



n+1



≤ lim inf

1/n

≤ lim sup

1/n

≤ lim sup



n+1



Radius and Interval of Convergence

The implied domain of the series in Example 2.6 turned out to be an interval (2, 6). Somewhat amaz-

ingly, the root test (Theorem 2.3) shows that the same is true for every power series!

Theorem 2.7 (Root Test for Power Series). Given a power series

∑

(x −c)

, deﬁne

R =

lim sup

1/n

The precisely one of the following statements holds:

R ∈ (0, ∞) the series converges absolutely when

x −c

< R and diverges when

x −c

> R

R = ∞ the series converges absolutely for all x ∈ R

R = 0 the series converges only at the center x = c

Proof. For each ﬁxed x ∈ R, let b

= a

(x −c)

and apply the root test to

∑

, noting that

lim sup

1/n











lim sup

1/n

x −c

if R ∈ (0, ∞)

0 if R = ∞ or x = c

∞ if R = 0 and x = c

In the ﬁrst situation, lim sup

1/n

< 1 ⇐⇒

x −c

< R, etc.

Deﬁnition 2.8. The radius of convergence is the value R deﬁned in Theorem 2.7. The interval of conver-

gence is the set of x ∈ R for which the series converges; its implied domain.

Radius of convergence Interval of convergence

R = 0, ∞ ( c − R, c + R), (c −R, c + R], [c − R, c + R) or [c − R, c + R]

∞ R = (−∞, ∞)

0 {c}

In the ﬁrst case, convergence/divergence at the endpoints of the interval of convergence must be

tested for separately.

The ratio test (Corollary 2.4) provides a more user-friendly version.

Corollary 2.9 (Ratio Test for Power Series). If the limit exists, R = lim

n→∞



n+1



The ratio test is weaker than the root test: as Example 2.10.5 shows, there exist series for which the

ratio test is be inconclusive.

Since

≥ 0, we here adopt the conventions

= ∞,

∞

= 0. With similar caveats, one can write R = lim inf

−1/n

Since every sequence has a limit superior, this really is a deﬁnition. Whether one can easily compute R is another matter. . .

Examples 2.10. 1. The series

∑

∞

n=1

is centered at 0. The ratio test tells us that

R = lim

n→∞



n+1



= lim

n→∞

1/n

1/(n + 1)

= lim

n→∞

n + 1

= 1

Test the endpoints of the interval of convergence separately:

x = 1

∑

= ∞ diverges

x = −1

∑

(−1)

converges (conditionally)

We conclude that the interval of convergence is [−1, 1).

It can be seen (later) that the series converges to −ln(1 − x)

on its interval of convergence. As in Example 2.6, this function

has a larger domain (−∞, −1), than that of the series.

−1

−3 −2 −1 1

y =

∞

∑

n=0

y = −ln(1 − x)

2. The series

∑

∞

n=1

similarly has

R = lim

n→∞



n+1



= lim

n→∞

(n + 1)

= 1

Since

∑

is absolutely convergent, we conclude that the power series also converges abso-

lutely at x = ±1; the interval of convergence is [−1, 1].

3. The series

∑

∞

n=0

converges absolutely for all x ∈ R, since

R = lim

n→∞



n+1



= lim

n→∞

(n + 1)!

= lim

n→∞

(n + 1) = ∞

You should recall from elementary calculus that this series converges to the natural exponential

function exp(x) = e

everywhere on R; indeed this is one of the common deﬁnitions of the

exponential function.

4. The series

∑

∞

n=0

n!x

has R = lim

(n+1)!

= 0. It therefore converges only at its center x = 0.

5. Let a





if n is even and





if n is odd. If we try to apply the ratio test to the series

∑

∞

n=0

, we see that



n+1



(





2n+1

if n even





2n+1

if n odd

=⇒ lim sup



n+1



= ∞ = 0 = lim inf



n+1



The ratio test is therefore inconclusive. However, by the root test,

1/n

(

if n even

if n odd

=⇒ R =

lim sup

1/n

3/2

It is easy to check that the series diverges at x = ±

; the interval of convergence is (−

With the help of the root test the domain of a power series is fully understood. Limits, continuity,

differentiability and integrability are more delicate. We will return to these once we’ve developed

some of the ideas around convergence for sequences of functions.

Exercises 2.23. Key concepts: Power Series, Radius/interval of convergence R =

lim sup

1/n

1. For each power series, ﬁnd the radius and interval of convergence:

(a)

∑

(−1)

(b)

∑

(n + 1)

(x −3)

(c)

∑

√

(d)

∑

√

(x + 7)

(e)

∑

(x −π)

(f)

∑

√

2n+1

2. For each n ∈ N let a



4+2(−1)



(a) Find lim sup

1/n

, lim inf

1/n

, lim sup



n+1



and lim inf



n+1



(b) Does the series

∑

converge? What about

∑

(−1)

? Why?

∑

3. Suppose that

∑

has radius of convergence R. If lim sup

> 0, prove that R ≤ 1.

4. On the interval (−

), express the series in Example 2.10.5 as a simple function.

(Hints: Use geometric series formulæ and the fact that the value of an absolutely convergent series is

independent of rearrangements)

5. Consider the power series

∞

∑

n=1

(x −7)

5n+1

(x −7) +

(x −7)

+ ···

Since only one in ﬁve of the terms are non-zero, it is a little tricky to analyze using a na

ıve

application of our standard tests.

(a) Explain why the ratio test for power series (Corollary 2.9) does not apply.

(b) Writing the series as

∑

(x −7)

, observe that







m−1

(m−1)

if m ≡ 1 mod 5

0 otherwise

Use the root test (Theorem 2.7) and your understanding of elementary limits to directly

compute the radius of convergence.

∑

(x −7)

5n+1

∑

. Apply the ratio test for inﬁnite series (Corol-

lary 2.4): what do you observe? Use your observation to compute the radius of conver-

gence of the original series in a simpler manner than part (a).

(d) Finally, check the endpoints to determine the interval of convergence.

2.24 Uniform Convergence

In this section we consider sequences ( f

) of functions f

: U → R and their limits.

Example 2.11. For each n ∈ N, deﬁne f

: (0, 1) → R : x 7→ x

. Several examples are graphed.

(x)

0 1

(x)

0 1

(x)

0 1

(x)

0 1

There are several useful notions of convergence for sequences of functions. The simplest is where,

for each input x, ( f

(x)) is treated as a distinct sequence of real numbers.

Deﬁnition 2.12. Suppose a function f and a sequence of functions ( f

) are given, all with domain

U. We say that ( f

) converges pointwise to f on U if,

∀x ∈ U, lim

n→∞

(x) = f (x)

It is common to write ‘ f

→ f pointwise.’ For reference, here are two equivalent rephrasings:

1. ∀x ∈ U, lim

n→∞

(x) − f (x)

= 0;

2. ∀x ∈ U, ∀ϵ > 0, ∃N such that n > N =⇒

(x) − f (x)

< ϵ.

As we’ll see shortly, the relative position of the quantiﬁers (∀x, ∃N) is crucial: in this deﬁnition, the

value of N is permitted to depend on x as well as ϵ.

Example (2.11, mk. II). The sequence ( f

) converges pointwise on the domain U = (0, 1) to

f : (0, 1) → R : x 7→ 0

As a sanity check, we prove this explicitly. First observe that

(x) − f (x)

= x

Suppose x ∈ (0, 1), that ϵ > 0 is given, and let N =

ln ϵ

ln x

Then

n > N =⇒ n ln x < ln ϵ =⇒ x

< ϵ

where the inequality switches sign since ln x < 0.

(x)

0 1

···

The example is nice in that a sequence of continuous functions converges pointwise to a continuous

function. Unfortunately, this desirable situation is not universal. . .

Example (2.11, mk. III). Deﬁne

: (0, 1] → R : x 7→ x

Each g

is a continuous function, however its pointwise limit

g(x) =

(

0 if x < 1

1 if x = 1

has a jump discontinuity at x = 1.

(x)

0 1

···

We’d like the limit of a sequence of continuous functions to itself be continuous. With this goal in

mind, we make a tighter deﬁnition.

Deﬁnition 2.13. ( f

) converges uniformly to f on U if either

1. sup

x∈U

(x) − f (x)

−−−→

n→∞

0, or,

2. ∀ϵ > 0, ∃N such that ∀x ∈ U, n > N =⇒

(x) − f (x)

< ϵ

A common notation is f

⇒ f , though we won’t use it.

2ǫ

f (x)

(x)

As pictured, whenever n > N, the graph of f

(x) must lie between f (x) ± ϵ.

We’ll show that statements 1 and 2 are equivalent momentarily. For the present, compare with the

corresponding statements for pointwise convergence:

• As with continuity versus uniform continuity, the distinction comes in the order of the quantiﬁers:

in uniform convergence, x is quantiﬁed after N and so the same N works for all locations x.

• Uniform convergence implies pointwise convergence.

Example (2.11, mk. IV). For the ﬁnal time we revisit our main example. If f

(x) = x

and f (x) = 0

are deﬁned on U = (0, 1), then f

→ f non-uniformly. We show this using both criteria.

1. For every n,

sup

x∈(0,1)

(x) − f (x)

= sup{x

: 0 < x < 1} = 1 −→ 0

which plainly fails to converge to zero.

2. Suppose the convergence were uniform and let ϵ =

. Then

∃N ∈ N such that ∀x ∈ (0, 1), n > N =⇒ x

Since N ∈ N, a simple choice results in a contradiction;

x =





N+1

∈ (0, 1) =⇒ x

N+1

−ǫ

Theorem 2.14. The criteria for uniform convergence in Deﬁnition 2.13 are equivalent.

Proof. (1 ⇒ 2) This follows immediately from the fact that

∀x ∈ U,

(x) − f (x)

≤ sup

x∈U

(x) − f (x)

(2 ⇒ 1) Suppose ϵ > 0 is given. Then

∃N ∈ R such that ∀x ∈ U, n > N =⇒

(x) − f (x)

But then

n > N =⇒ sup

x∈U

(x) − f (x)

≤

< ϵ

Amazingly, this subtle change of deﬁnition is enough to preserve continuity.

Theorem 2.15. Suppose ( f

) is a sequence of continuous functions. If f

→ f uniformly, then f is

continuous.

Proof. We demonstrate the continuity of f at a ∈ U. Let ϵ > 0 be given.

• Since f

→ f uniformly,

∃N such that ∀x ∈ U, n > N =⇒

f (x) − f

(x)

• Choose any n > N. Since f

is continuous at a,

∃δ > 0 such that

x −a

< δ =⇒

(x) − f

(a)

(†)

Put these together with the triangle inequality to see that

x −a

< δ =⇒

f (x) − f (a)

≤

f (x) − f

(x)

(x) − f

(a)

(a) − f (a)

= ϵ

We need not have ﬁxed a at the start of the proof. Rewriting (†) to become

∃δ > 0 such that ∀x, a ∈ U,

x −a

< δ =⇒

(x) − f

(a)

proves a related result.

Corollary 2.16. Suppose ( f

) is a sequence of uniformly continuous functions. If f

→ f uniformly,

then f is also uniformly continuous.

Examples 2.17. 1. Let f

(x) = x +

. This is continuous on R for all x, and converges pointwise

to the continuous function f : x 7→ x.

(a) On any bounded interval [−M, M] the convergence f

→ f is uniform,

sup

x∈[−M,M]

(x) − f (x)

= sup



: x ∈ [−M, M]



−−−→

n→∞

(b) On any unbounded interval, R say, the convergence is non-uniform,

sup

x∈R

(x) − f (x)

= sup



: x ∈ R



= ∞

2. Consider f

(x) =

1+x

; this is continuous on

(−1, ∞) and converges pointwise to

f (x) =











0 if x > 1

if x = 1

1 if −1 < x < 1

We consider the convergence f

→ f on several

intervals.

(x)

−1 0 1 2 3

(a) On [2, ∞), the pointwise limit is continuous. Moreover, f

(x) is decreasing, whence

sup

x∈[2,∞)

(x) −0

1 + 2

−−−→

n→∞

and the convergence is uniform. Alternatively; if ϵ ∈ ( 0, 1), let N = log

(ϵ

−1

−1), then

∀x ≥ 2, n > N =⇒

(x) −0

1 + x

≤

1 + 2

= ϵ

The same argument shows that f

→ f uniformly on any interval [a, ∞) where a > 1.

(b) On [1, ∞) the convergence is not uniform, since the pointwise limit is discontinuous,

f (x) =

(

0 if x > 1

if x = 1

sup

x∈[1,∞)

(x) − f (x)

= sup



1 + x

: x > 1



−−−→

n→∞

(d) Similarly, for any a ∈ (0, 1), the convergence f

→ f is uniform on [0, a], this time to the

(continuous) constant function f (x) = 1,

sup

x∈[0,a]

(x) −1



1 −

1 + a



1 + a

−−−→

n→∞

(e) Finally, on (−1, 1) the convergence is not uniform,

sup

x∈[0,1)

(x) − f (x)

= sup



1 + x

: x ∈ [0, 1)



−−−→

n→∞

Exercises 2.24. Key concepts: Pointwise & Uniform Convergence, Uniform conv preserves continuity

1. For each sequence of functions deﬁned on [0, ∞):

(i) Find the pointwise limit f (x) as n → ∞.

(ii) Determine whether f

→ f uniformly on [0, 1].

(iii) Determine whether f

→ f uniformly on [1, ∞).

(a) f

(x) =

(b) f

(x) =

1 + x

(x) =

n + x

(d) f

(x) =

1 + nx

(e) f

(x) =

1 + nx

2. Let f

(x) =



x −



. If f (x) = x

, we clearly have f

→ f pointwise on any domain.

(a) Prove that the convergence is uniform on [−1, 1].

(b) Prove that the convergence is non-uniform on R.

3. For each sequence, ﬁnd the pointwise limit and decide if the convergence is uniform.

(a) f

(x) =

1+2 cos

(nx)

√

for x ∈ R.

(b) f

(x) = cos

(x) on [−π/2, π/2].

4. For each n ∈ N, consider the continuous function

: [0, 1] → R : x 7→ nx

(1 − x)

(a) Given 0 ≤ x < 1, let a ∈ (x, 1). Explain why ∃N such that

n > N =⇒

n+1

(x)

≤ a

(x)

Hence conclude that the pointwise limit of ( f

) is the zero function.

(b) Use elementary calculus ( f

′

(x) = 0 ⇐⇒ . . .) to prove that the maximum value of f

located at x

1+n

. Hence compute

sup

x∈[0,1]

(x) − f (x)

and use it to show that the convergence f

→ 0 is non-uniform.

This shows that the converse to Theorem 2.15 is false, even on a bounded interval: the continuous

sequence ( f

) converges non-uniformly to a continuous function. Sketches of several f

are below.

0 1

−1

y = f

(x)

0 1

−1

y = f

(x)

0 1

−1

y = f

(x)

0 1

−1

y = f

(x)

5. Explain where the proof of Theorem 2.15 fails if f

→ f non-uniformly.

2.25 More on Uniform Convergence

While we haven’t yet developed calculus, our familiarity with basic differentiation and integration

makes it natural to pause to consider the interaction of these concepts with sequences of functions.

We also consider a Cauchy-criterion for uniform convergence, which leads to the useful Weierstraß

M-test.

Example 2.18. Recall that f

(x) = x

converges uniformly to f (x) = 0 on any interval [0, a] where

a < 1. We easily check that

(x) dx =

n + 1

n+1

−−−→

n→∞

0 =

f (x) dx

In fact the sequence of derivatives converge here also

(x) = nx

n−1

−−−→

n→∞

0 = f

′

(x)

It is perhaps surprising that integration interacts more nicely with uniform limits than does differen-

tiation. We therefore consider integration ﬁrst.

Theorem 2.19. Let f

→ f uniformly on [a, b] where the functions f

are integrable. Then f is

integrable on [a, b] and

lim

n→∞

(x)dx =

f (x)dx

Proof. Given ϵ > 0, note that

2(b−a)

dx =

. Since f

→ f uniformly, ∃N such that

∀x ∈ [a, b] , n > N =⇒

(x) − f (x)

2(b − a)

=⇒ f

(x) −

2(b − a)

< f (x) < f

(x) +

2(b − a)

=⇒

(x) dx −

≤

f (x) dx ≤

(x) dx +

=⇒



(x) dx −

f (x) dx



≤

< ϵ

The appearance of uniform convergence in the proof is subtle. If N = N(ϵ) were allowed to depend

on x, then the integral

(x) dx would be meaningless: Which n would we consider? Larger than

N(x, ϵ) for which x? Taking n ‘larger’ than all the N(x, ϵ) might produce the absurdity n = ∞!

This assumes f is already integrable. Once we’ve properly deﬁned (Riemann/Darboux) integrability at the end of the

course, we can insert the following

(x) dx −

≤ L( f ) ≤ U( f ) ≤

(x) dx +

=⇒ 0 ≤ U( f ) − L( f ) ≤ ϵ =⇒ U( f ) = L( f )

where U( f ) and L( f ) are the upper and lower Darboux integrals of f ; equality shows that f is integrable on [a, b].

Examples 2.20. 1. Uniform convergence is not required for the integrals to converge as we’d like. For

instance, recall that extending the previous example to the domain [0, 1] results in non-uniform

convergence; however, we still have

(x) dx =

n + 1

−−−→

n→∞

0 =

f (x) dx

2. To obtain a sequence of functions f

→ f for which

↛

f requires a bit of creativity.

Consider the sequence

: [−1, 1] → R : x 7→

(

n − n

x if 0 < x <

0 otherwise

If 0 < x < 1, then for large n ∈ N we have

x ≥

=⇒ f

(x) = 0

−1 1

(x)

We conclude that f

→ 0 pointwise. Since the area under f

is a triangle with base

and height

n, the integral is constant and non-zero;

−1

(x) dx =

= 0 =

−1

f (x) dx

It should be obvious that the convergence f

→ 0 is non-uniform; why?

Derivatives and Uniform Limits We’ve already seen that a uniform limit of differentiable functions

might be differentiable (Example 2.18). As the next example shows, this should not be expected in

general, since even uniform limits of differentiable functions can have corners.

Example 2.21. For each n ∈ N, consider the function

: R → R : x 7→

(

≥

• Each f

is differentiable: f

′

(x) =











1 if x ≥

nx if

−1 if x ≤ −

• f

converges pointwise to f (x) =

, which is non-

differentiable at x = 0.

• f

→ f uniformly since

sup

x∈[−1,1]

(x) − f (x)

= f

(0) =

→ 0

(x)

−1 0 1

−1

′

(x)

−1 1

−

If our goal is to transfer differentiability to the limit of a sequence of functions, then we have some

work to do.

Theorem 2.22. Suppose ( f

) is a sequence and f , g functions, all with domain [a, b]. Suppose also:

• f

→ f pointwise;

• Each f

is differentiable with continuous derivative;

• f

′

→ g uniformly.

Then f

→ f uniformly on [a, b] and f is differentiable with derivative g.

The issue in the previous example is that the pointwise limit of the derived sequence ( f

′

) is discontin-

uous at x = 0 and therefore f

′

→ g isn’t uniform!

Proof. For any x ∈ [a, b] , the fundamental theorem of calculus (part II) tells us that

′

(t) dt = f

(x) − f

(a)

As n → ∞, Theorem 2.19 says the left side converges to

g(t) dt and the right to f (x) − f (a) (both

pointwise). Since f

′

→ g uniformly, we see that g is continuous and can apply the fundamental

theorem (part I):

g(t) dt = f (x) − f (a) is differentiable with derivative g.

The uniformity of the convergence f

→ f follows from Exercise 10.

Uniformly Cauchy Sequences and the Weierstraß M-Test

Recall that one may use Cauchy sequences to demonstrate convergence without knowing the limit in

advance. An analogous discussion is available for sequences of functions.

Deﬁnition 2.23. A sequence of functions ( f

) is uniformly Cauchy on U if

∀ϵ > 0, ∃N ∈ N such that ∀x ∈ U, m, n > N =⇒

(x) − f

(x)

< ϵ

Example 2.24. Let f

(x) =

∑

k=1

sin k

x be deﬁned on R. Given ϵ > 0, let N =

, then

m > n > N =⇒

(x) − f

(x)



∑

k=n+1

sin k



≤

∑

k=n+1

≤

∑

k=n+1

k(k −1)

∑

k=n+1

k − 1

−

= ϵ

whence ( f

) is uniformly Cauchy.

Without this continuity assumption, the fundamental theorem of calculus doesn’t apply and the proof requires an

alternative approach. One can also weaken the hypotheses: if f

′

→ g uniformly and ( f

(x)) converges for at least one

x ∈ [a, b], then there exists f such that f

→ f is uniform and f

′

= g.

As with sequences of real numbers, uniformly Cauchy sequences converge; in fact uniformly!

Theorem 2.25. A sequence ( f

) is uniformly Cauchy on U if and only if it converges uniformly to

some f : U → R.

Proof. (⇒) Let ( f

) be uniformly Cauchy on U. For each x ∈ U, the sequence ( f

(x)) ⊆ R is Cauchy

and thus convergent. Deﬁne f : U → R to be the pointwise limit:

f (x) := lim

n→∞

(x)

We claim that f

→ f uniformly. Let ϵ > 0 be given, then ∃N ∈ N such that

m > n > N =⇒

(x) − f

(x)

=⇒ f

(x) −

< f

(x) < f

(x) +

=⇒ f

(x) −

≤ f (x) ≤ f

(x) +

(take limits as m → ∞)

=⇒

(x) − f (x)

≤

< ϵ

(⇐) This is Exercise 2.

Example (2.24, mk. II). Since ( f

) is uniformly Cauchy on R, it converges uniformly to some f :

R → R. It seems reasonable to write

f (x) =

∞

∑

n=1

sin n

The graph of this function looks somewhat bizarre:

−1

f (x)

2ππ−2π −π

Since each f

is (uniformly) continuous, Theorem 2.15 says that f is also (uniformly) continuous. By

Theorem 2.19, f (x) is integrable, indeed

f (x) dx = lim

n→∞

∑

k=1

−k

−4

cos k



∞

∑

n=1

(cos n

a −cos n

which converges (comparison test) for all a, b. By contrast, the derived sequence

′

(x) =

∑

k=1

cos k

does not converge for any x since lim

n→∞

cos n

x = 0. We should therefore expect (though we offer no

proof) that f is nowhere differentiable.

The example generalizes. Suppose (g

) is a sequence of functions on U and deﬁne the series

∑

(x)

as the pointwise limit of the sequence ( f

) of partial sums

∞

∑

k=k

(x) := lim

n→∞

(x) where f

(x) =

∑

k=k

(x)

whenever the limit exists. The series is said to converge uniformly whenever ( f

) does so. Theorems

2.15, 2.19 and 2.22 immediately translate.

Corollary 2.26. Let

∑

be a series of functions converging uniformly on U. Then:

1. If each g

is (uniformly) continuous then

∑

is (uniformly) continuous.

2. If each g

is integrable, then

∑

(x) dx =

∑

(x) dx.

3. If each g

is continuously differentiable, and the sequence of derived partial sums f

′

converges

uniformly, then

∑

is differentiable and

∑

(x) =

∑

′

(x).

As an application of the uniform Cauchy criterion, we obtain an easy test for uniform convergence.

Theorem 2.27 (Weierstraß M-test). Suppose (g

) is a sequence of functions on U. Moreover assume:

1. (M

) is a non-negative sequence such that

∑

converges.

2. Each g

is bounded by M

; that is

(x)

≤ M

Then

∑

(x) converges uniformly on U.

Proof. Let f

(x) =

∑

k=k

(x) deﬁne the sequence of partial sums. Since

∑

converges, its sequence

of partial sums is Cauchy (the Cauchy criterion for inﬁnite series); given ϵ > 0,

∃N such that m > n > N =⇒

∑

k=n+1

< ϵ

However, by assumption,

m > n > N =⇒

(x) − f

(x)



∑

k=n+1

(x)



≤

∑

k=n+1

(x)

≤

∑

k=n+1

< ϵ

The sequence of partial sums is uniformly Cauchy and thus uniformly convergent.

Example 2.28. Given the series

∞

∑

n=1

1+cos

(nx)

sin(nx), we clearly have



1 + cos

(nx)

sin(nx)



≤

for all x ∈ R

Since

∑

converges, the M-test shows that the original series converges uniformly on R.

Exercises 2.25. Key concepts: Uniform convergence preseves integration, Uniform Cauchyness, M-test

1. For each n ∈ N, let f

(x) = nx

when x ∈ [0, 1) and f

(1) = 0.

(a) Prove that f

→ 0 pointwise on [0, 1].

(Hint: recall Exercise 2.24.4 if you’re not sure how to prove this)

(b) By considering the integrals

(x) dx show that f

→ 0 is not uniform.

2. Prove that if f

→ f uniformly, then the sequence ( f

) is uniformly Cauchy.

3. (a) Suppose ( f

) is a sequence of bounded functions on U and suppose that f

→ f converges

uniformly on U. Prove that f is bounded on U.

(b) Give an example of a sequence of bounded functions ( f

) converging pointwise to f on

[0, ∞), but for which f is unbounded.

4. The sequence deﬁned by f

(x) =

1+nx

(Exercise 2.24.1) converges uniformly on any closed

interval [a, b] where 0 < a < b.

(a) Check explicitly that

(x) dx →

f (x) dx, where f = lim f

(b) Is the same thing true for derivatives?

5. Let f

(x) = n

−1

sin n

x be deﬁned on R.

(a) Prove that f

converges uniformly on R.

(b) Check that

(t) dt converges for any x ∈ R.

′

) converge? Explain.

6. Use the M-test to prove that

∞

∑

n=1

deﬁnes a continuous function on [−1, 1].

7. Prove that

∞

∑

n=1

sin x

(n+1)

converges uniformly to a continuous function on the interval [−2, 2] .

8. Prove that if

∑

converges uniformly on a set U and if h is a bounded function on U, then

∑

converges uniformly on U.

(Warning: you cannot simply write

∑

= h

∑

)

9. Consider Example 2.20.2.

(a) Check explicitly that the convergence isn’t uniform by computing sup

x∈[−1,1]

(x) − f (x)

(b) Prove that f

→ 0 pointwise on (0, 1] using the ϵ–N deﬁnition of convergence: that is,

given ϵ > 0 and x ∈ (0, 1], ﬁnd an explicit N(x, ϵ) such that

n > N =⇒

f (x)

< ϵ

What happens to your choice of N(x, ϵ) as x → 0

10. Suppose ( f

′

) converges uniformly on [a, b] and that each f

′

is continuous.

(a) Use the fact that ( f

′

) is uniformly Cauchy to prove that ( f

) is uniformly Cauchy and thus

converges uniformly to some function f .

(Hint:

(x) − f

(x)



′

(t) − f

′

(t) dt



. . .)

(b) Explain why we need not have assumed the existence of f in Theorem 2.22.

2.26 Differentiation and Integration of Power Series

We now specialize our recent results to power series. While everything will be stated for series

centered at x = 0, all are easily translated to arbitrary centers.

Theorem 2.29. Let

∑

be a power series with radius of convergence R > 0 and let T ∈ (0, R).

Then:

1. The series converges uniformly on [−T, T].

2. The series is uniformly continuous on [−T, T] and continuous on (−R, R).

Proof. This is a straightforward application of the Weierstraß M-test (Theorem 2.27). For each k,

deﬁne M

, and observe that

T < R =⇒

∑

converges absolutely =⇒

∑

converges

By the M-test and Corollary 2.26, the power series converges uniformly on [−T, T] to a uniformly

continuous function.

Finally, every x ∈ (−R, R) lies in some such interval (take T =

), whence the power series is

continuous on (−R, R).

Example 2.30. On its interval of convergence (−1, 1), the geometric series

∞

∑

n=0

converges pointwise

1−x

; convergence is uniform on any interval [−T, T] ⊆ (−1, 1).

We needn’t use the Theorem for this is simple to verify directly: writing f , f

for the series and its

partial sums,

(x) − f (x)



1 − x

n+1

1 − x

−

1 − x



n+1

1 − x



=⇒ sup

x∈[−T,T]

(x) − f (x)

n+1

1 − T

−−−→

n→∞

By contrast, the convergence is non-uniform on (−1, 1), since

sup

x∈(−1,1)

(x) − f (x)

= ∞

Theorem 2.31. Suppose a power series

∑

has radius of convergence R > 0. Then the series is

integrable and differentiable term-by-term on the interval (−R, R). Indeed for any x ∈ (−R, R),

∞

∑

n=0

∞

∑

n=1

n−1

and

∞

∑

n=0

dt =

∞

∑

n=0

n + 1

n+1

where both series also have radius of convergence R.

Proof. Let f (x) =

∑

have radius of convergence R, and observe that

lim sup

1/n

= lim n

1/n

lim sup

1/n

whence

∑

also has radius of convergence R. At any given non-zero x ∈ (−R, R), we may write

∞

∑

n=1

n−1

= x

−1

∞

∑

n=1

to see that the derived series also has radius of convergence R. On any interval [−T, T] ⊆ (−R, R), the

derived series converges uniformly (Theorem 2.29). Since each a

is continuously differentiable,

Corollary 2.26 says that f is differentiable on [−T, T] and that

′

(x) =

∞

∑

n=0

∞

∑

n=1

n−1

Since any x ∈ ( −R, R) lies in some such interval [−T, T], we are done.

Exercise 7 discusses the corresponding result for integration.

We postpone the canonical examples until after the next result.

Continuity at Endpoints?

There is one small hole in our analysis. A series

∑

with radius of convergence R converges and

is continuous on (−R, R). But what if it also converges at x = ±R? Is the series continuous at the

endpoints? The answer is yes, though demonstrating this small beneﬁt requires a lot of work!

Theorem 2.32 (Abel’s Theorem). Power series are continuous on their full interval of convergence.

Examples 2.33. 1. Apply our results to the geometric series;

(1 − x)

1 − x

∞

∑

n=1

n−1

∞

∑

n=0

(n + 1)x

= 1 + 2x + 3x

+ 4x

+ ···

ln( 1 −x) = −

1 −t

dt = −

∞

∑

n=0

n + 1

n+1

= −

∞

∑

n=1

= −



x +

+ ···



where both are valid on (−1, 1). In fact the ﬁrst series has exactly this interval of convergence,

whereas the second has [−1, 1). By Abel’s Theorem and the fact that logarithms are continuous,

we have equality at x = −1 and recover the famous identity

ln 2 =

∞

∑

n=1

(−1)

n+1

= 1 −

−

+ ···

This also shows that while integrated and differentiated series have the same radius of conver-

gence as the original, convergence at the endpoints need not be the same.

2. Substitute x 7→ −x

in the geometric series and integrate term-by-term: if

< 1, then

1 + x

∞

∑

n=0

(−1)

=⇒ arctan x =

∞

∑

n=0

(−1)

2n + 1

2n+1

In fact the arctangent series also converges at x = ±1; Abel’s Theorem says it is continuous on

[−1, 1]. Since arctangent is continuous (on R!) we recover another famous identity

= arctan 1 =

∞

∑

n=0

(−1)

2n + 1

= 1 −

−

+ ···

As with the identity for ln 2, this is a very slowly converging alternating series and therefore

doesn’t provide an efﬁcient method for approximating π.

3. The series f (x) =

∞

∑

n=0

(−1)

(2n)!

has radius of convergence ∞. Differentiate to obtain

′

(x) =

∞

∑

n=1

(−1)

2n−1

(2n −1)!

∞

∑

n=0

(−1)

n+1

(2n + 1)!

2n+1

This series is also valid for all x ∈ R. Differentiating again,

′′

(x) =

∞

∑

n=0

(−1)

n+1

(2n)!

= −

∞

∑

n=0

(−1)

(2n)!

= −f (x)

Recalling that f (x) = cos x is the unique solution to the initial value problem

(

′′

(x) = −f (x)

f (0) = 1, f

′

(0) = 0

We conclude that, ∀x ∈ R,

cos x =

∞

∑

n=0

(−1)

(2n)!

sin x = −f

′

(x) =

∞

∑

n=0

(−1)

(2n + 1)!

2n+1

These last expressions can be taken as the deﬁnitions of sine and cosine. As promised earlier, con-

tinuity and differentiability of these functions now come for free! The only real downside of this

deﬁnition is believing that it has anything to do with right-triangles!

We can similarly deﬁne other common transcendental functions using power series: for instance

exp(x) =

∞

∑

n=0

Example 2.33.1 could also be taken as a deﬁnition of the logarithm on the interval ( 0, 2],

ln x = ln(1 − (1 − x)) = −

∞

∑

n=1

(1 − x)

∞

∑

n=1

(−1)

n+1

(x −1)

though this is unnecessary since it is more natural to deﬁne ln as the inverse of the exponential.

Proof of Abel’s Theorem (non-examinable)

This requires a lot of work, so feel free to omit on a ﬁrst reading!

First observe that there is nothing to check unless 0 < R < ∞. By the change of variable x 7→ ±

, it

is enough for us to prove the following:

∞

∑

n=0

convergent and f (x) =

∞

∑

n=0

on (−1, 1) =⇒ lim

x→1

−

f (x) =

∞

∑

n=0

Proof. Let s

∑

k=0

and write s = lim s

∑

. It is an easy exercise to check that

∑

k=0

= s

+ (1 − x)

n−1

∑

k=0

< 1, then (since s

→ s) lim s

= 0, whence we obtain

∀x ∈ (−1, 1), f (x) = (1 − x)

∞

∑

n=0

Let ϵ ∈ (0, 1) be given and ﬁx x ∈ (0, 1). Then

∃N ∈ N such that n > N =⇒

−s

(∗)

Use the geometric series formula

∞

∑

n=0

1−x

and write h(x) = (1 − x)



∑

n=0

−s)x



to observe

f (x) −s



(1 − x)

∞

∑

n=0

−s



(1 − x)

∞

∑

n=0

−s(1 −x)

∞

∑

n=0



(1 − x)

∞

∑

n=0

−s)x



= (1 − x)



∑

n=0

−s)x

∞

∑

n=N+1

−s)x



≤ (1 − x)



∑

n=0

−s)x



+ (1 − x)



∞

∑

n=N+1

−s)x



(△-inequality)

< h(x) +

(1 − x)



∞

∑

n=N+1



(by (∗))

≤ h(x) +

Since h > 0 is continuous and h(1) = 0, ∃δ > 0 such that x ∈ (1 − δ, 1) =⇒ h(x) <

(the computa-

tion of a suitable δ is another exercise).

We conclude that lim

x→1

−

f (x) = s.

Exercises 2.26. Key concepts: Power series continuous on ( −R, R), uniformly on any [−T, T] ⊂ (−R, R),

Power series differentiable and integrable term-by-term on (−R, R)

1. (a) Prove that

∑

∞

n=1

(1−x)

for

< 1.

(b) Evaluate

∑

∞

n=1

∑

∞

n=1

and

∑

∞

n=1

(−1)

2. (a) Starting with a power series centered at x = 0, evaluate the integral

1/2

1+x

dx as an

inﬁnite series.

(b) (Harder) Repeat part (a) but for

1+x

dx. What extra ingredients do you need?

3. The probability that a standard normally distributed random variable X lies in the interval [a, b]

is given by the integral

P(a ≤ X ≤ b) =

√

2π

exp



−



Find P(−1 ≤ X ≤ 1) as an inﬁnite series.

4. If f (x) =

∑

∞

n=0

(−1)

(2n)!

is deﬁned as in Example 2.33.3, prove that



f (x)





′

(x)



= 1.

What does (the converse of) Pythagoras’ Theorem say about f (x), at least when both it and

′

(x) are positive?

(Hint: Differentiate and evaluate at zero!)

5. Deﬁne c(x) =

∑

∞

n=0

(2n)!

and s(x) =

∑

∞

n=0

2n+1

(2n+1)!

(a) Prove that c

′

(x) = s(x) and that s

′

(x) = c(x).

(b) Prove that c(x)

−s(x)

= 1 for all x ∈ R.

(These functions are the hyperbolic sine and cosine: s(x) = sinh x and c(x) = cosh x)

6. Let a, b ∈ (−1, 1). Extending Example 2.30, show that the convergence

∑

1−x

is non-

uniform on any interval of the form (−1, a) or (b, 1).

7. Prove the integration part of Theorem 2.31.

8. Prove or disprove: If a series converges absolutely at the endpoints of its interval of convergence

then its convergence is uniform on the entire interval.

9. Complete the proof of Abel’s Theorem:

(a) Let s

∑

k=0

be the partial sum of the series

∑

. For each n, prove that,

∑

k=0

= s

+ (1 − x)

n−1

∑

k=0

(b) Suppose x > 0. Let S = max{

−s

: n ≤ N} and prove that h(x) ≤ S(1 − x

N+1

). Hence

ﬁnd an explicit δ that completes the ﬁnal step.

2.27 The Weierstraß Approximation Theorem

A major theme of analysis is approximation; for instance power series are an example of (uniform)

approximation by polynomials. It is reasonable to ask whether any function can be so approximated.

In 1885, Weierstraß answered a speciﬁc case in the afﬁrmative.

Theorem 2.34 (Weierstraß). If f : [a, b] → R is continuous, then there exists a sequence of polyno-

mials converging uniformly to f on [a, b].

Suitable polynomials can be deﬁned in various ways. By scaling the domain, it is enough to do this

on [a, b] = [0, 1] where perhaps the simplest approach is via the Bernstein Polynomials,

f (x) :=

∑

k=0









(1 − x)

n−k

(

)

k!(n−k)!

is the binomial coefﬁcient)

We omit the proof due to length; Weierstraß’ original argument was completely different. Instead we

compute a couple of examples and give an important interpretation/application.

Examples 2.35. 1. Suppose f (x) = 2x if x <

and f (x) = 1 otherwise.

f (x) = f ( 0)(1 − x) f (0) + f (1)x = x

f (x) = f ( 0)(1 − x)

+ 2 f (

)x(1 − x) + f (1)x

= 2x(1 − x) + x

= x(2 − x)

0 1

f (x) = f ( 0)(1 − x)

+ 3 f (

)x(1 − x)

+ 3 f (

(1 − x) + f (1)x

= 0(1 − x)

+ 2x(1 − x)

+ 3x

(1 − x) + x

= x(2 − x) = B

f (x)

f (x) = 0(1 − x)

+ 2x(1 − x)

+ 6x

(1 − x)

+ 4x

(1 − x) + x

= x(x

−2x

+ 2)

The Bernstein polynomials B

f (x), B

f (x) and B

f (x) are drawn.

2. Now assume f (x) = x if x <

and f (x) = 1 − x otherwise.

f (x) = f ( 0)(1 − x) + f (1)x = 0

f (x) = x(1 − x)

f (x) = 0(1 − x)

+ x(1 − x)

+ x

(1 − x) + 0x

= x(1 − x) = B

f (x)

0 1

f (x) = f ( 0)(1 − x)

+ f (

)·4x(1 −x)

+ f (

)·6x

(1 − x)

+ f (

)·4x

(1 − x) + f (1)x

= x(1 − x)

+ 3x

(1 − x)

+ x

(1 − x)

= x(1 − x)(1 + x − x

)

B´ezier curves (just for fun!)

The Bernstein polynomials arise naturally when con-

sidering B´ezier curves. These have many applications,

particularly in computer graphics. Given three points

A, B, C, deﬁne points on the line segments

−→

AB and

−→

for each t ∈ [0, 1], via

−→

AB(t) = (1 −t)A + tB

−→

BC(t) = (1 −t)B + tC

These points move at a constant speed along the cor-

responding segments. Now consider a point on the

moving segment between the points deﬁned above:

0 1

−→

R(t) := (1 − t)

−→

AB(t) + t

−→

BC(t) = (1 − t)

A + 2t(1 − t)B + t

This is the quadratic B´ezier curve with control points A, B, C. The 2

Bernstein polynomial for a function

f is simply the quadratic B

ezier curve with control points

(

0, f (0)

)



, f (

)



and

(

1, f (1)

)

. The

picture

above shows B

f (x) for the above example.

We can repeat the construction with more control points: with four points A, B, C, D, one constructs

−→

AB(t),

−→

BC(t),

−→

CD(t), then the second-order points between these, and ﬁnally the cubic B

ezier curve

R(t) : = (1 − t)



(1 − t)

−→

AB(t) + t

−→

BC(t)



+ t



(1 − t)

−→

BC(t) + t

−→

CD(t)



= (1 − t)

A + 3t(1 − t)

B + 3t

(1 − t)C + t

where we now recognize the relationship to the 3

Bernstein polynomial.

The pictures show cubic B

ezier curves: the ﬁrst is the graph of the Bernstein polynomial

f (x) = 0(1 − x)

+ 3x(1 − x)

+ 3x

(1 − x) +

while the second is for the four given control points A, B, C, D.

Click on any of the pictures to see all of them move.

Exercises 2.27. Key concepts: Every continuous function is the uniform limit of a polynomial sequence

1. Show that the closed bounded interval assumption in the approximation theorem is required

by giving an example of a continuous function f : (−1, 1) → R which is not the uniform limit

of a sequence of polynomials.

2. If g : [a, b] → R is continuous, then f (x) := g



(b − a)x + a



is continuous on [0, 1]. If P

→ f

uniformly on [0, 1], prove that Q

→ g uniformly on [a, b], where

(x) = P



x −a

b − a



3. Use the binomial theorem to check that every Bernstein polynomial for f (x) = x is B

f (x) = x

itself!

4. Find a parametrization of the cubic B

ezier curve with control points (1, 0), (0, 1), (−1, 0) and

(0, −1). Now sketch the curve.

(Use a computer algebra package if you like!)

5. (Hard) Show that the Bernstein polynomials for f (x) = x

are given by

f (x) =

x +

n −1

and thus verify explicitly that B

f → f uniformly.

3 Differentiation

Differentiation grew out of the problem of instantaneous velocity. Velocity can only easily be measured

as an average over a time interval:

if an object travels ∆d meters in ∆t seconds, then its average

velocity is v

∆d

∆t

−1

. An early ‘deﬁnition’ (dating to the 1300s) makes the instantaneous velocity

equal to the constant velocity that would be observed if a body were to stop accelerating: while

useless for the purposes of measurement, this is essentially Newton’s ﬁrst law regarding inertial

motion (1687). We also see the concept of the tangent line beginning to appear. Indeed if one graphs

position against time, intuition tells us:

• The graph of inertial (constant speed) motion is a straight line whose slope is the velocity.

• The tangent line to a curve has slope equal to the instantaneous velocity.

The problem of ﬁnding, deﬁning and computing instantaneous velocity thus morphed into the con-

sideration of tangent lines to curves. With the advent of analytic geometry in the early 1600s, math-

ematicians such as Fermat and Descartes pioneered versions of the familiar secant (‘cutting’) line

method for computing tangents.

∆t

∆d

v =

∆d

∆t

Instantaneous velocity equals constant

velocity corresponding to tangent line

Secant lines approximate tangent line as t → a

The average velocity of the particle over the time interval [a, t] is the slope of the secant line, namely

(a, t) =

d(t) − d(a)

t − a

Since the secant lines approximate the tangent line as t approaches a, it seems reasonable that we

should compute the instantaneous velocity in this manner:

v(a) = lim

t→a

(a, t) = lim

t→a

d(t) − d(a)

t − a

This is, of course, the modern deﬁnition of the derivative.

Even a modern technique such as Doppler-shift compares measurements separated by the extremely small period of a

light or soundwave. These are still therefore average velocities, albeit taken over very small time intervals.

3.28 Basic Properties of the Derivative

Deﬁnition 3.1. Let f : U → R and a ∈ U

◦

an interior point. We say that f is differentiable at a if the

following limit exists (is ﬁnite!)

lim

x→a

f (x) − f (a)

x −a

We call this limit the derivative of f at a and denote its value by

d f



x=a

or f

′

(a).

If f

′

(a) exists for all a ∈ U then f is differentiable (on U); the derivative becomes a function f

′

(x) =

d f

The two notations are partly attributable to the primary founders of calculus: Issac Newton and

Gottfried Leibniz. Each has its pros and cons and you should be comfortable with both.

One-sided derivatives Differentiability only makes sense at interior points of U since the deﬁning

limit is two-sided. Left- and right-derivatives may be deﬁned using one-sided limits; differentiability

is then equivalent to these being equal. All results in this section hold for one-sided derivatives with

suitable (sometimes tedious) modiﬁcations. It is common, though strictly incorrect, to say that f is

differentiable on [a, b) if it is differentiable on the interior (a, b) and right-differentiable at a. In these

notes we will strictly adhere to Deﬁnition 3.1: differentiable means two-sided.

Examples 3.2. Basic examples should be familiar from elementary calculus.

1. Let f (x) = x

+ 4x. Then, for any a ∈ R,

lim

x→a

f (x) − f (a)

x −a

= lim

x→a

+ 4x −a

−4a

x −a

= lim

x→a

(x − a)(x + a + 4)

x −a

= lim

x→a

(x + a + 4) = 2a + 4

Note how the deﬁnition of lim

x→a

allows us to cancel the x − a terms from the numerator and

denominator. We conclude that f is differentiable (on R) and that f

′

(x) = 2x + 4.

2. Let g(x) =

x+1

2x−3

. Then, for any a =

lim

x→a

f (x) − f (a)

x −a

= lim

x→a

x −a



x + 1

2x −3

−

a + 1

2a −3



= lim

x→a

5a −5x

(x − a)(2x −3)(2a −3)

= lim

x→a

−5

(2x −3)(2a −3)

−5

(2a −3)

f is therefore differentiable on its domain R \{

} with derivative f

′

(x) =

−5

(2x−3)

The familiar expressions

′

(a) = lim

h→0

f (a + h) − f (a)

, f

′

(x) = lim

h→0

f (x + h) − f (x)

are equivalent to the original deﬁnition (Exercise 5). While seemingly simpler, they sometimes lead

to nastier calculations: see what happens if you try the previous example in this language. . .

We now turn to perhaps the most well-known result of elementary calculus.

Theorem 3.3 (Power Law). Let r ∈ R. Then f (x) = x

is differentiable with f

′

(x) = rx

r−1

The domains of f and f

′

depend messily on r, but the formula holds at least on the interval (0, ∞).

We leave a complete proof to the exercises and instead consider a few generalizable examples.

Examples 3.4. 1. If n ∈ N and a ∈ R, a simple factorization yields

lim

x→a

− a

x −a

= lim

x→a

(x − a)(x

n−1

+ ax

n−2

+ ···+ a

n−2

x + a

n−1

)

x −a

(∗)

= lim

x→a

n−1

+ ax

n−2

+ ···+ a

n−2

x + a

n−1

) = na

n−1

We conclude that

= nx

n−1

2. If f (x) = x

−1

and a = 0, then

lim

x→a

−1

− a

−1

x −a

= lim

x→a

a − x

ax(x −a)

= lim

x→a

−1

= −

from which we conclude that f

′

(x) = −x

−2

A similar approach followed by the factorization (∗) proves the

power law for all negative integer exponents:

−n

− a

−n

x −a

− x

(x − a)

= ···

−2

−1

−2 −1 1 2

3. To differentiate x

1/n

, substitute x = y

and observe case 1. For

instance, if g(x) = x

1/3

and a = 0, then y = x

1/3

and b = a

1/3

yield

lim

x→a

1/3

− a

1/3

x −a

= lim

y→b

y − b

−b

−2/3

=⇒ g

′

(x) =

−2/3

Note that g is not differentiable at x = 0!

−1

−2 −1 1 2

We could similarly compute the derivative for all rational exponents, though it is much easier to wait

for the chain rule. The power law for irrational exponents is somewhat more ticklish.

Corollary 3.5 (Basic Transcendental Functions). Recalling our development of power series in

Chapter 2, the power law (for positive integers!) is all we need to see that

exp(x) = exp(x),

sin x = cos x,

cos x = −sin x

It is also possible to develop these results independently of power series (see e.g. Exercise 12).

Failure of differentiability

It is instructive to consider how a function might fail to be differentiable. Firstly, a familiar fact shows

that functions are not differentiable at discontinuities.

Lemma 3.6. If f is differentiable at a then f is continuous at a.

Proof. Just take the limit (think carefully why this works!):

lim

x→a

f (x) = lim

x→a



f (x) − f (a)

x −a

(x − a) + f (a)



= f

′

(a)(0 −0) + f (a) = f (a)

It remains to consider situations when a function is continuous but not differentiable.

Examples 3.7. The following exemplify all situations where a function is continuous on an interval

and differentiable everywhere except at a single interior point. As with isolated discontinuities, these

are classiﬁed by considering the three ways in which the derivative limit might not converge.

1. A vertical tangent line occurs when the limit is inﬁnite. For instance, g(x) = x

1/3

at x = 0.

2. Corners occur when the one-sided limits are unequal (could be inﬁnite). For instance, f (x) =

is not differentiable at zero, with one-sided limits

lim

x→0

−

x −0

= lim

x→0

= 1 = lim

x→0

−

x −0

= lim

x→0

−

−x

= −1

Indeed f is differentiable everywhere except at zero, with

′

(x) =

(

1 if x > 0

−1 if x < 0

A cusp describes the special case where the one-sided limits are ∞ = −∞.

3. A singularity is where left- and/or right-limits do not exist.

The standard example is

f (x) =

(

x sin

if x = 0

0 if x = 0

which is continuous on R and differentiable everywhere ex-

cept at zero: the details are in Exercise 10.

−

Singularities and vertical tangent lines can also prevent one-sided differentiability.

More esoteric examples of non-differentiability are possible:

• Utilizing series, we can create functions which are continuous on an interval but nowhere differ-

entiable! For an example, see Exercise 15.

• It is also possible to construct a function which differentiable (and thus continuous) at precisely

one point; can you think of an example?

The Basic Rules of Differentiation

Theorem 3.8. Let f , g be differentiable and k, l be constants.

1. (Linearity) The function k f + lg is differentiable with (k f + lg)

′

= k f

′

+ lg

′

2. (Product rule) The function f g is differentiable with ( f g)

′

= f

′

g + f g

′

3. (Inverse functions) Suppose f is bijective, b = f

−1

(a) is an interior point of dom f

−1

, and

′

(a) = 0, then f

−1

is differentiable at b and



y=b

−1

(y) =

′

(a)

′



−1

(b)



Proof. Parts 1 and 2 follow from the limit laws:

lim

x→a

(k f + lg)(x) − (k f + lg)(a)

x −a

= lim

x→a



f (x) − f (a)

x −a

+ l

g(x) − g(a)

x −a



= k f

′

(a) + lg

′

(a)

lim

x→a

f (x)g(x) − f (a)g(a)

x −a

= lim

x→a



f (x) − f (a)

x −a

g(x) + f (a)

g(x) − g(a)

x −a



= f

′

(a)g(a) + f (a)g

′

(a)

Note where we used the continuity of g in the second line (lim g(x) = g(a)). Part 3 is an exercise.

The inverse function rule should be intuitive: since the graphs of f and f

−1

are related by reﬂection

in the diagonal y = x, gradients at corresponding points are reciprocals. The result feels even more

natural in Leibniz’s notation:

dy/dx

Examples 3.9. 1. Linearity permits the differentiation of any polynomial: e.g.,



+ 13x



= 7

+ 13

= 14x + 52x

2. The product rule extends the reach of differentiation to include simple combinations: e.g.,

sin x) =





sin x + x

sin x = 4x

sin x − x

cos x

3. Inverse trigonometric functions can now be differentiated: e.g.,

y = sin

−1

x =⇒

sin

−1

x =





−1

cos y

1 −sin

√

1 − x

4. Deﬁne the natural logarithm to be the inverse of the (bijective!) exponential function exp(x):

y = ln x ⇐⇒ x = exp y

It follows that

ln x =





−1

exp y

The full details, and the justiﬁcation that exp x = e

, are in Exercise 14.

Theorem 3.10 (Chain Rule). If g is differentiable at a, and f is differentiable at g(a), then f ◦ g is

differentiable at a with derivative

( f ◦ g)

′

(a) = f

′



g(a)



′

(a)

In Leibniz’s notation,

d( f ◦g)

d f

: this looks like a simple cancellation of the dg terms. . .

Proof. Since f and g are differentiable, a is interior to dom(g) and g(a) is interior to dom( f ). Since g

is continuous at a, there must exist some open interval U ∋ a for which x ∈ U =⇒ g(x) ∈ dom( f ).

Deﬁne γ : dom( f ) → R via

γ(v) =











−f



g(a)



v−g(a)

if v = g(a)

′



g(a)



if v = g(a)

(∗)

Since f is differentiable at g(a),we see that γ is continuous there: indeed lim

v→g(a)

γ(v) = f

′



g(a)



For any x ∈ U \ {a}, let v = g(x) in (∗). Then



g(x)



− f



g(a)



x −a

= γ



g(x)



g(x) − g(a)

x −a

Take limits as x → a for the result.

Corollary 3.11 (Quotient Rule). Suppose f and g are differentiable. Then

is differentiable when-

ever g(x) = 0. Moreover





′

g − f g

′

The proof is an exercise.

Examples 3.12. 1. By the quotient rule,

tan x =

sin x

cos x

cos

x + sin

cos

= sec

2. We can now differentiate highly involved combinations of elementary functions:



tan(e

) −

sin x



= 8xe

sec

) −

7 sin x −7x cos x

sin

This is completely unjustiﬁed since dg does not (for us) have independent meaning. The same problem appears in a

famously ﬂawed one-line ‘proof’ of the chain rule:

lim

x→a



g(x)



− f



g(a)



x −a

= lim

x→a



g(x)



− f



g(a)



g(x) − g(a)

lim

x→a

g(x) − g(a)

x −a

The second limit doesn’t make sense unless g(x) = g(a) for all x on some punctured neighborhood of a: in particular, g(x)

cannot be constant! The faulty argument may be repaired by replacing this difference quotient with f

′



g(a)



whenever

g(x) = g(a), before taking the limit. This is precisely what γ



g(x)



does in the correct proof.

Exercises 3.28. Key concepts: Differentiability, Basic rules: linearity, power, product, chain, quotient

1. Use Deﬁnition 3.1 to calculate the derivatives.

(a) f (x) = x

at x = 2 (b) g(x) = x + 2 at x = a

cos x at x = 0 (d) r(x) =

3x+4

2x−1

at x = 1

2. Differentiate the function f (x) = cos



−3x



using the chain and product rules.

3. (a) Prove the quotient rule (Corollary 3.11) by combining the chain and product rules.

(b) Prove the inverse derivative rule (Theorem 3.8, part 3).

(Hint: You can’t simply differentiate 1 =



−1

(x)



using the chain rule; why not?)

4. (a) Find the derivatives of secant, cosecant and cotangent using the quotient rule.

(b) Why did we choose the positive square-root when computing

sin

−1

x? What is the

standard domain of arcsine, and what happens at x = ±1?

5. Using the deﬁnition of the derivative, and supposing that f is differentiable at a, prove that

′

(a) = lim

h→0

f (a + h) − f (a)

= lim

h→0

f (a + h) − f (a −h)

6. Use induction to prove the power law

= nx

n−1

when n ∈ N using only the product rule

and the fact that

x = 1.

7. Prove that f (x) = x

is differentiable everywhere and compute its derivative.

8. Show that f (x) = x

2/3

has a cusp (see Example 3.7.2) at x = 0.

9. Show that following function is differentiable everywhere and compute its derivative:

f (x) =

(

sin

if x = 0

0 if x = 0

Moreover, prove that the derivative f

′

is discontinuous at x = 0.

10. Prove that the function in Example 3.7.3 is differentiable everywhere except at x = 0.

11. Suppose f (x) = x

whenever x ∈ Q and f (x) = 0 whenever x ∈ Q. At what values of x is f

differentiable? Prove your assertion.

12. (a) Suppose 0 < h <

. Use the picture to show that

0 <

1 −cos h

< sin

and sin h < h < tan h

Hence conclude that lim

h→0

sin h

= 1 and lim

h→0

1−cos h

= 0.

(b) Use part (a) to prove that

sin x = cos x

cos h

sin h

tan h

13. (Hard) Use induction to prove the Leibniz rule (general product rule):

( f g)

(n)

∑

k=0





(k)

(n−k)

Warning! The last two exercises are much longer and & tougher: have a go if you appreciate a

challenge.

14. The Exponential Function & the Power Law

The ratio tests shows that the power series exp(x) :=

∑

∞

n=0

converges for all real x. Deﬁne

e := exp(1). Certainly e

makes sense whenever x ∈ Q. If x is irrational, instead deﬁne

:= sup{e

: q ∈ Q, q < x}

The goal of this question is to prove that exp(x) = e

. As a nice bonus we recover Bernoulli’s

limit identity e = lim

n→∞



1 +



and obtain a complete proof of the power law!

(a) For all x, y ∈ R, prove that exp(x + y) = exp(x) exp(y)

(Hint: use the binomial theorem and change the order of summation)

(b) Show that exp(x) is always positive, even when x < 0.

(Hint: x ≥ 0 =⇒ exp(x) ≥ 1 + x; take limits then apply part (a))

(d) Prove that e

= exp(x). Do this in three stages:

• If x ∈ N, use part (a). Now check for x ∈ Z

−

• If x =

∈ Q, ﬁrst compute



exp(

)



• If x is irrational, consider a sequence of rational numbers q

< x with e

→ e

. . .

(e) Let ln : (0, ∞) → R be the inverse function of exp. Prove the logarithm laws:

ln(xy) = ln x + ln y and ln x

= r ln x

(Just do this when r ∈ N; in general, another argument like part (d) is required)

(f) We’ve already seen that

ln y =

. Use the fact that

ln y = lim

h→0

ln(y + h) −ln y

to prove that exp(x) = lim

n→∞



1 +



, thus recovering Bernoulli’s deﬁnition of e.

(g) For any r ∈ R, deﬁne x

:= exp(r ln x). Hence obtain the power law for any exponent.

15. A Very Strange Function

Here is a classic example of a continuous but nowhere-differentiable function!

Let f be the sawtooth function deﬁned by f (x) =

whenever x ∈ [−1, 1] and extending

periodically to R so that f (x + 2) = f (x). Now deﬁne g : R → R via

g(x) =

∞

∑

n=0





f (4

−2 −1 0 1 2

f (x) and iterations to n = 3 g(x) (really n = 6, but can you tell?!)

(a) Prove that g is well-deﬁned and continuous on R.

(b) Let x ∈ R and m ∈ N be ﬁxed. Deﬁne h

= ±

·4

−m

where the sign is chosen so that no

integers lie strictly between 4

x and 4

(x + h

) = 4

x ±

For each n ∈ N

, deﬁne



(x + h

)



− f (4

Prove the following

≤ 4

with equality when n = m.

ii. n > m =⇒ k

= 0.

(Hint:

f (y) − f (z)

≤

y − z

: when is this an equality?)



g(x + h

) − g(x)



≥

+ 1)

Hence conclude that g is nowhere differentiable.

3.29 The Mean Value Theorem

A key result in elementary calculus, this should be very familiar from your previous studies.

Theorem 3.13 (Mean Value Theorem/MVT). Let f be continuous on [a, b] and differentiable on

(a, b). Then there exists ξ ∈ (a, b) such that f

′

(ξ) =

f (b)−f (a)

b−a

This follows easily from two lemmas.

Lemma 3.14. 1. (Critical Points) Suppose g is bounded on (a, b) and attains its maximum or mini-

mum at ξ ∈ (a, b). If g is differentiable at ξ then g

′

(ξ) = 0.

2. (Rolle’s Theorem) Suppose g is continuous on [a, b], differentiable on (a, b), and g(a) = g(b).

Then there exists ξ ∈ (a, b) such that g

′

(ξ) = 0.

The main result is obtained by subtracting a straight line and applying Rolle’s theorem to

g(x) = f (x) −

f (b) − f (a)

b − a

(x − a)

and observing that g(a) = f (a) = g(b) and g

′

(x) = f

′

(x) −

f (b)−f (a)

b−a

g(x)

a b

Critical Points/Rolle’s Theorem

f (x)

a b

Mean Value Theorem

In the pictures, the orange and green lines are parallel: the average slope over the interval [a, b] equals

the gradient/derivative f

′

(ξ).

Proof of Lemma. 1. Suppose ξ ∈ (a, b) is a maximum: that is, g(x) ≤ g(ξ) for all x = ξ. Then

g(x) − g(ξ)

x −ξ

(

≤ 0 whenever x > ξ

≥ 0 whenever x < ξ

Now take the one-sided limits: since g is differentiable at ξ, we see that

0 ≤ lim

x→ξ

g(x) − g(ξ)

x −ξ

= g

′

(ξ) = lim

x→ξ

−

g(x) − g(ξ)

x −ξ

≤ 0

Otherwise said g

′

(ξ) = 0. The case when ξ is a minimum is similar.

2. By the Extreme Value Theorem (1.11), g is bounded and attains its bounds. If the extrema both

occur at the endpoints a, b, then g is constant: any ξ ∈ (a, b) satisﬁes the result. Otherwise, at

least one extreme occurs at some ξ ∈ (a, b): part 1 says that g

′

(ξ) = 0.

Examples 3.15. 1. Let f (x) = (x −1)

(4 − x) + x on [a, b] = [1, 4]: this is roughly the above picture

illustrating the mean value theorem. Compute the average slope and the derivative,

f (b) − f (a)

b − a

= 1, f

′

(x) = 2(x −1)(4 − x) − (x − 1)

+ 1 = −3x

+ 12x −8

and observe that

′

(ξ) =

f (b) − f (a)

b − a

⇐⇒ 3ξ

−12ξ + 9 = 0 ⇐⇒ ξ = 1 or 3

Since only 3 lies in the interval ( 1, 4), this is the value ξ satisfying the mean value theorem.

2. We ﬁnd the maximum and minimum values of g(x) = x

−14x

+ 24x on the interval [0, 2].

The function is differentiable, with

′

(x) = 4x

−28x + 24 = 4(x −2)(x −1)(x + 3)

By the Lemma, the locations of the extrema are either the end-

points x = 0, 2 or locations with zero derivative (x = 1). Since

f (0) = 0, f (1) = 11, f (2) = 8

we conclude that max( f ) = f (1) = 11 and min( f ) = f (0) = 0.

g(x)

0 1 2

Consequences of the Mean Value Theorem Several simple corollaries relate to monotonicity.

Deﬁnition 3.16. Suppose f : I → R is deﬁned on an interval I. We say that f is:

Increasing (monotone-up) on I if x < y =⇒ f (x) ≤ f (y)

Decreasing (monotone-down) on I if x < y =⇒ f (x) ≥ f (y)

We say strictly increasing/decreasing if the inequalities are strict.

Examples 3.17. 1. f : x 7→ x

is strictly increasing on [0, ∞)

and strictly decreasing on (−∞, 0].

2. The ﬂoor function f : x 7→ ⌊x⌋(the greatest integer less

than or equal to x) is increasing, but not strictly, on R.

−2

−1

g(x)

−2 −1 1 2 3

Corollary 3.18. Suppose f is differentiable on an interval I. Then

1. f

′

≥ 0 on I ⇐⇒ f is increasing on I

2. f

′

≤ 0 on I ⇐⇒ f is decreasing on I

3. f

′

= 0 on I ⇐⇒ f is constant on I

Proof. (Part 1, ⇒) Let x < y where x, y ∈ I. By the mean value theorem, ∃ξ ∈ (x, y) such that

f (y) − f (x)

y − x

= f

′

(ξ) whence f

′

(ξ) ≥ 0 =⇒ f (y) ≥ f (x)

(⇐) For the converse, use the deﬁnition of derivative: f

′

(ξ) = lim

x→ξ

f (x)−f (ξ)

x−ξ

. If f is increasing, then

x > ξ =⇒ f (x) ≥ f ( ξ) =⇒ f

′

(ξ) ≥ 0

Parts 2 and 3 are similar.

More care is required when relating f

′

> 0 to f being strictly increasing (see Exercise 5). The corollary

also yields a couple of (hopefully familiar) ﬂashbacks to elementary calculus.

Corollary 3.19. Let I be an open interval.

1. (Anti-derivatives on an interval) If f

′

(x) = g

′

(x) on I, then ∃c such that g(x) = f (x) + c on I.

2. (First derivative test) Suppose f is continuous on I and differentiable except perhaps at ξ. If

(

′

(x) < 0 whenever x < ξ, and

′

(x) > 0 whenever x > ξ

then f has its minimum value at x = ξ

The statement for a maximum is similar.

Examples 3.20. 1. Since

sin( 3x

+ x) = (6x + 1) cos(3x

+ x) on (the interval) R, whence all

anti-derivatives of f (x) = (6x + 1) cos(3x

+ x) are given by

f (x) dx =

(6x + 1) cos(3x

+ x) dx = sin(3x

+ x) + c

As is typical in calculus, we use the indeﬁnite integral notation

f (x) dx for anti-derivatives.

2. If f (x) = x

2/3

x/3

, then f

′

(x) =

−1/3

(2 + x)e

x/3

By Lemma 3.14, the only possible critical points are at

x = 0 or −2. The sign of the derivative is also clear:

−2 0

′

(x) > 0 f

′

(x) < 0 f

′

(x) > 0

f (x)

−3 −2 −1 0 1

By the 1

derivative test, f has a maximum at x = −2 and a minimum at x = 0.

We ﬁnish this section by tying together the mean and intermediate value theorems.

Theorem 3.21 (IVT for Derivatives). Suppose f is differentiable on an interval I containing a < b,

and that L lies between f

′

(a) and f

′

(b). Then ∃ξ ∈ (a, b) such that f

′

(ξ) = L.

If f

′

(x) is continuous, this is just the intermediate value theorem applied to f

′

; surprisingly, continuity

of f

′

is not required. A full proof is in Exercise 7.

Exercises 3.29. Key concepts: Differentiability, Basic rules: linearity, power, product, chain, quotient

1. Determine whether the conclusion of the mean value theorem holds for each function on the

given interval. If so, ﬁnd a suitable point ξ. If not, state which hypothesis fails.

(a) x

on [−1, 2] (b) sin x on [0, π] (c)

on [−1, 2]

(d) 1/x on [−1, 1] (e) 1/x on [1, 3]

2. Suppose f and g are differentiable on an interval I containing a < b and that f (a) = f (b) = 0.

By considering h(x) = f (x)e

g(x)

, prove that f

′

(ξ) + f (ξ)g

′

(ξ) = 0 for some ξ ∈ (a, b).

3. (a) Use the Mean Value Theorem to prove that x < tan x for all x ∈ (0,

(b) Prove that

sin x

is strictly increasing on (0,

sin x for all x ∈ [0,

4. Suppose that

f (x) − f (y)

≤ (x −y)

for all x, y ∈ R. Prove that f is a constant function.

5. (a) Prove that f

′

> 0 on an interval I =⇒ f is strictly increasing on I.

(b) Show that the converse of part (a) is false.

6. If f is differentiable on an interval I such that f

′

(x) = 0 for all x ∈ I, use the intermediate value

theorem for derivatives to prove that f is either strictly increasing or strictly decreasing.

7. (Intermediate value theorem for derivatives) Let f , a, b and L be as in Theorem 3.21, deﬁne

g : I → R by g(x) = f (x) − Lx, and let ξ ∈ [a, b] be such that

g(ξ) = min



g(x) : x ∈ [a, b]



(a) Why can we be sure that ξ exists? If ξ ∈ (a, b), explain why f

′

(ξ) = L.

(b) Assume WLOG that f

′

(a) < f

′

(b). Prove that g

′

(a) < 0 < g

′

(b). By considering

lim

x→a

g(x)−g(a)

x−a

, show that ∃x > a for which g(x) < g(a). Hence complete the proof.

8. Suppose f

′

exists on (a, b), and is continuous except for a discontinuity at c ∈ (a, b).

(a) Suppose lim

x→c

′

(x) = L < f

′

(c). By taking ϵ =

′

(c)−L

in the deﬁnition of this limit

and applying IVT for derivatives, obtain a contradiction.

Hence argue that c cannot be a removable or a jump discontinuity.

(b) Similarly, show that f

′

cannot have an inﬁnite discontinuity by considering lim

x→c

′

(x) = ∞.

′

can have an essential discontinuity. Recall

(Exercise 3.28.9) that

f : R → R : x 7→

(

sin

x = 0

0 x = 0

is differentiable on R, but has discontinuous derivative at x = 0.

i. Use x

2nπ

and y

(2n+1)π

to show that f

′

has an essential discontinuity at x = 0.

ii. Prove that if lim s

= 0 and lim f

′

) = M, then M ∈ [ −1, 1].

iii. Prove that for any L ∈ [−1, 1], there is a sequence (t

) for which lim f

′

) = L.

(Hint: Use IVT for derivatives)

3.30 L’Hˆopital’s Rule

We are often required to consider indeterminate forms: limits which do not yield easily to the standard

limits laws. For instance, while it is tempting to write

lim

x→0

sin 2x

−1

lim sin 2x

lim e

−1

(∗)

this is an incorrect application of the limit laws since the resulting quotient has no meaning.

Deﬁnition 3.22. An indeterminate form is any limit where a na

ıve application of the limit laws results

in a meaningless expression: the primary types are

∞

, ∞ −∞, 0 · ∞, 0

, 0

∞

, and 1

∞

Examples 3.23. 1. lim

x→7

(x −7)

x−7

is an indeterminate form of type 0

∞

2. Our motivating example (∗) may correctly be evaluated using the deﬁnition of the derivative:

lim

x→0

sin 2x

−1

= lim

x→0

sin 2x − 0

x −0

−1





x=0

sin 2x





x=0



−1

By considering lim

x→0

3a sin 2x

2(e

−1)

, we see that an indeterminate form of type

can take any value a!

The approach generalizes, if non-rigorously: if f , g are differentiable at a and f (a) = 0 = g(a), then

lim

x→a

f (x)

g(x)

= lim

x→a

f (x) − f (a)

x −a

g(x) − g(a)

′

(a)

′

(a)

Our goal is to fully justify this result and extend to several situations:

• One-sided limits, including when a = ±∞ .

• When lim f (x) = 0 exists, but f (a) does not (g(x), g(a) similarly).

• Indeterminate forms of type

∞

(lim f (x) = ∞, etc.).

• When the RHS cannot be cleanly evaluated: for instance g

′

(a) = 0 or if the original limit is ±∞.

Here is the full result.

Theorem 3.24 (L’H ˆopital’s Rule). Let a ∈ R ∪{±∞} and suppose functions f and g satisfy:

1. lim

x→a

′

(x)

′

(x)

= L for some L ∈ R ∪ {±∞}, and,

2. (a) lim

x→a

f (x) = lim

x→a

g(x) = 0, or (b) lim

x→a

g(x) = ∞ (no condition on f )

Then lim

x→a

f (x)

g(x)

= L. The same result holds for one-sided limits.

The full proof is a behemoth—we postpone this until after several examples. In part because of this,

and because examples can often be evaluated more instructively using elementary methods (as in the

above example), l’H

opital’s rule is often discouraged in elementary calculus.

Examples 3.25. 1. If f (x) = e

and g(x) = 21x −17, then lim

x→∞

f (x)

g(x)

has type

∞

. By l’H

opital’s rule,

lim

x→∞

′

(x)

′

(x)

= lim

x→∞

= ∞ =⇒ lim

x→∞

21x −17

= ∞

2. For an example of type

, consider f (x) = x

−9 and g(x) = ln(4 −x):

lim

x→3

−

′

(x)

′

(x)

= lim

x→3

−

−1/(4 − x)

= lim

x→3

−

2x(x −4) = −6 =⇒ lim

x→3

−

−9

ln( 4 −x)

= −6

3. One can apply the rule repeatedly: for example

lim

x→0

−1 −4x

= lim

x→0

−4

= lim

x→0

16e

= 8

This is a generally accepted abuse of protocol: one shouldn’t really state the ﬁrst limit until one

knows the last limit exists! As long as everything works, you are ﬁne. However. . .

4. It is crucially important that the limit lim

′

exists before applying l’H

opital’s rule! Consider

f (x) = x + cos x and g(x) = x: certainly lim

x→∞

f (x)

g(x)

has type

∞

, however

lim

x→∞

′

(x)

′

(x)

= lim

x→∞

1 −sin x

does not exist! In this case the rule is unnecessary: appealing to the squeeze theorem,

f (x)

g(x)

= 1 +

cos x

−−−→

x→∞

5. For another reason for why l’H

opital’s rule is often prohibited in Freshman calculus, consider

lim

x→0

sin x

= lim

x→0

cos x

= 1

This appears legitimate. However, recall (Exercise 3.28.12) that this limit is used to demonstrate

sin x = cos x; to use this to calculate the limit on which it depends is circular logic!

The remaining indeterminate forms (Deﬁnition 3.22) may be modiﬁed so that l’H

opital’s rule applies.

Examples 3.26. 1. An indeterminate form of type ∞ −∞ may be transformed to one of type

before

applying the rule (twice):

lim

x→0

−1

−

= lim

x→0

x + 1 − e

x(e

−1)

(type

)

= lim

x→0

1 −e

−1 + xe

(still type

)

= lim

x→0

−e

+ xe

= −

2. For an indeterminate form of type 1

∞

, we use the log laws & continuity of the exponential:

lim

x→0

(1 + sin x)

1/x

= exp



lim

x→0

ln( 1 + sin x)



(type

)

= exp



lim

x→0

cos x

1 + sin x



= e

Proving l’Hˆopital’s Rule

The complete argument is very lengthy. It starts with an extension of the Mean Value Theorem.

Lemma 3.27 (Extended Mean Value Theorem). Fix a < b, suppose f , g are continuous on [a, b] and

differentiable on (a, b). Then there exists ξ ∈ (a, b) such that



f (b) − f (a)



′

(ξ) =



g(b) − g(a)



′

(ξ)

Proof. Apply the standard mean value theorem (really Rolle’s theorem) to

h(t) =



f (b) − f (a)



g(t) −



g(b) − g(a



) f (t)

which satisﬁes h(a) = h(b).

Now for the main event. If you do nothing else, read the following proof of the simplest case. Every-

thing else is a modiﬁcation.

Proof (Case (a)/type

, with right limits). Suppose we have a form of type

= lim

x→a

f (x)

g(x)

taking right-

limits at a ﬁnite location a, and that the resulting limit L is ﬁnite.

First observe that condition 1 forces the existence of an interval (a, b) on which f , g are differentiable

and g

′

(x) = 0. Everything follows from the deﬁnition the limit in condition 1, and Lemma 3.27:

Given ϵ > 0, ∃δ ∈ (0, b − a) such that a < ξ < a + δ =⇒



′

(ξ)

′

(ξ)

− L



(∗)

a < y < x < a + δ =⇒ ∃ξ ∈ ( y, x) such that

f (x) − f (y)

g(x) − g(y)

′

(ξ)

′

(ξ)

(†)

Since g

′

= 0, the usual mean value theorem says

∃c ∈ (y, x) such that g(x) − g(y) = g

′

(c)(x − y) = 0

whence we never divide by zero in (†). Combining (∗) and (†) , observe that

a < x < a + δ =⇒



f (x)

g(x)

− L



2(a)

= lim

y→a



f (x) − f (y)

g(x) − g(y)

− L



(†)

= lim

y→a



′

(ξ)

′

(ξ)

− L



(∗)

≤

< ϵ

Note that a < y < ξ(x, y) < x is a function of x, y here! Since ϵ > 0 is arbitrary, this is the required

result.

A complete proof for all indeterminate forms of type

follows from some simple modiﬁcations.

If a = −∞ : Replace the blue part of (∗) as follows:

Given ϵ > 0, ∃m ≤ b such that ξ < m =⇒



′

(ξ)

′

(ξ)

− L



The rest of the proof goes through after replacing a with −∞ and a + δ with m.

If L = ∞: Replace the green parts of (∗) with Given M > 0 and

′

(ξ)

′

(ξ)

> 2M. Fixing the rest of the

proof is again straightforward.

If L = −∞: Replace the green parts of (∗) with Given M > 0 and

′

(ξ)

′

(ξ)

< −2M.

Left-limits: If f , g are differentiable on (c, a), then the blue part may be replaced with either:

• (a ﬁnite) ∃δ ∈ (0, a − c) such that a −δ < ξ < a

• (a = ∞) ∃m ≥ c such that ξ > m

The blue and green parts of (∗) may be replaced independently.

Proof (Case (b), lim g(x) = ∞). This requires a little more care.

Since g

′

= 0, and lim

x→a

g(x) = ∞,

Exercise 3.29.6 says that g is strictly decreasing on (a, b). By replacing b by some

b ∈ (a, b), if necessary,

we may assume that

a < y < x < b =⇒ 0 < g(x) < g(y) (‡)

Assume a and L are ﬁnite and obtain (∗) and (†) as before. Let x ∈ (a, a + δ) be ﬁxed and multiply

(†) by

g(y)−g(x)

g(y)

(this is positive by (‡)): a little algebra and the triangle inequality tell us that

a < y < x =⇒

f (y)

g(y)

′

(ξ)

′

(ξ)

f (x)

g(y)

−

g(x)

g(y)

′

(ξ)

′

(ξ)

=⇒



f (y)

g(y)

− L



≤



′

(ξ)

′

(ξ)

− L



g(y)



f (x)

g(x)



L +



Since lim

y→a

g(y) = ∞ and x is ﬁxed, we see that there exists η ≤ x − a < δ such that

y ∈ (a, a + η) =⇒

g(y)



f (x)

g(x)



L +



Finally combine with (∗): given ϵ > 0, ∃η > 0 such that y ∈ (a, a + η) =⇒



f (y)

g(y)

− L



< ϵ.

The same modiﬁcations listed above complete the proof.

Forms of type

∞

? Instead of assumption 2. (b), why not simply assume lim f = lim g = ∞ and write

1/g

1/ f

to obtain

a form of type

? The problem is that the derivative of the ‘new’ denominator

−f

′

need not be non-zero on any

interval (a, b) and so condition 1. need not hold. We could modify this, but it would make for a weaker theorem. Example

3.25.4 illustrates the issue: f

′

(x) = 1 + sin x has zeros on any unbounded interval.

After the 2. (b) case is proved and we know that lim

= L, it is then clear that lim f must also be inﬁnite (unless L = 0 in

which case lim f could be anything and need not exist). This situation therefore really does deal with forms of type

∞

Exercises 3.30. Key concepts: Types of indeterminate forms, Formal statement of l’Hˆopital’s rule

1. Evaluate the limits, if they exist:

(a) lim

x→0

sin x − x

(b) lim

x→

−

tan x −

π −2x

x→0

(cos x)

1/x

(d) lim

x→0

(1 + 2x)

1/x

(e) lim

x→∞

+ x)

1/x

2. Suppose f is differentiable on (c, ∞) and that lim

x→∞

[ f (x) + f

′

(x)] = L is ﬁnite.

(a) Prove that lim

x→∞

f (x) = L and that lim

x→∞

′

(x) = 0.

(Hint: write f (x) =

f (x)e

)

(b) Does anything change if L exists and is inﬁnite?

3. If p

(x) is a polynomial of degree n, use induction to prove that lim

x→∞

(x)e

−x

= 0

4. Let f (x) = x + sin x cos x, g(x) = e

sin x

f (x) and h(x) =

2 cos x

sin x

( f (x) + 2 cos x)

(a) Prove that lim

x→∞

f (x) = ∞ = lim

x→∞

g(x) but that lim

x→∞

f (x)

g(x)

does not exist.

(b) If cos x = 0, and x is large, show that

′

(x)

′

(x)

= h(x).

x→∞

h(x) = 0. Explain why this does not contradict part (a)!

3.31 Taylor’s Theorem

A primary goal of power series is the approximation of functions. With this in mind, there are two

natural questions to ask of a function f :

1. Given c ∈ dom( f ), is there a series

∑

(x −c)

which equals f (x) on an interval containing c?

2. If we take the ﬁrst n terms of such a series, how accurate is this polynomial approximation?

Example 3.28. Recall the geometric series

f (x) =

1 − x

∞

∑

n=0

whenever − 1 < x < 1

The polynomial approximation

(x) =

∑

k=0

= 1 + x + ···+ x

1 − x

n+1

1 − x

has error

(x) = f (x) − p

(x) =

n+1

1 − x

−1 0 1

−

(x) = 1 + x + x

+ x

If x is close to 0, this is likely very small; for instance if x ∈



−



, then

(x)

≤

1 −





n+1

= 2

−n

However, when x is close to 1 the error is unbounded!

The above behavior occurs in general: the truncated polynomials provide better approximations

nearer the center of the series. To see this, we ﬁrst need to consider higher-order derivatives.

Deﬁnition 3.29. We write f

′′

for the second derivative of f , namely the derivative of its derivative

′′

(a) = lim

x→a

′

(x) − f

′

(a)

x −a

The existence of f

′′

(a) presupposes that f

′

exists on an (open) interval containing a. We can similarly

consider third, fourth, and higher-order derivatives. As a function, the n

derivative is written

(n)

(x) =

By convention, the zeroth derivative is the function itself f

(0)

(x) = f (x). We say that f is n times

differentiable at a if f

(n)

(a) exists, and inﬁnitely differentiable (or smooth) if derivatives of all orders exist.

Example 3.30. f (x) = x

is twice differentiable, with f

′′

(x) = 6

. It is smooth everywhere

except at x = 0, where third (and higher-order) derivatives do not exist.

Deﬁnition 3.31. Suppose f is n times differentiable at x = c. The n

Taylor polynomial p

of f

centered at c is

(x) :=

∑

k=0

(k)

(c)

(x −c)

= f (c) + f

′

(c)(x − c) +

′′

(c)

(x −c)

+ ···+

(n)

(c)

(x −c)

The remainder R

(x) is the error in the polynomial approximation

(x) = f (x) − p

(x) = f (x) −

∑

j=0

(k)

(c)

(x −c)

If f is inﬁnitely differentiable at x = c, then its Taylor series centered at x = c is the power series

f (x) =

∞

∑

n=0

(n)

(c)

(x −c)

When c = 0 this is known as a Maclaurin series.

For simplicity we’ll mostly work with Maclaurin series, with general situation hopefully being clear.

Examples 3.32. 1. If f (x) = e

, then f

(n)

(x) = 3

, from which the Maclaurin series is

f (x) =

∞

∑

n=0

2. If g(x) = sin 7x, then the sequence of derivatives is

7 cos 7x, −7

sin 7x, −7

cos 7x, 7

sin 7x, 7

cos 7x, −7

sin 7x, . . .

At x = 0, every even derivative is zero whereas the odd derivatives alternate in sign. The

Maclaurin series is easily seen to be

g(x) =

∞

∑

n=0

(−1)

2n+1

(2n + 1)!

2n+1

3. If h(x) =

√

x, then h

′

(x) =

−1/2

, h

′′

(x) =

−1

−3/2

, and h

′′′

(x) =

−5/2

, from which the

third Taylor polynomial centered at c = 1 is

(x) = h(1) + h

′

(1)(x −1) +

′′

(1)

(x −1)

′′′

(1)

(x −1)

= 1 +

(x −1) −

(x −1)

Rather than computing further examples, we ﬁrst develop a little theory that makes verifying Taylor

series much easier.

Named for Englishman Brook Taylor (1685–1731) and Scotsman Colin Maclaurin (1698–1746). Taylor’s general method

expanded on examples discovered by James Gregory and Issac Newton in the mid-to-late 1600s.

Differentiation of Taylor Polynomials and Series

Suppose P(x) =

∑

is a power series with radius of convergence R > 0. As we saw previously

(Theorem 2.31), P(x) is differentiable term-by-term on (−R, R). Indeed,

′

(x) =

∞

∑

j=1

j−1

=⇒ P

′

(0) = a

′′

(x) =

∞

∑

j=2

j(j −1)x

j−2

=⇒ P

′′

(0) = 2a

′′′

(x) =

∞

∑

j=3

j(j −1)(j −2)x

j−3

=⇒ P

′′′

(0) = 3!a

(k)

(x) =

∞

∑

j=k

j(j −1) ···(j −k + 1)x

j−k

∞

∑

j=k

j!a

(j − k)!

j−k

=⇒ P

(k)

(0) = k!a

Otherwise said, P is its own Maclaurin series! The same discussion holds for polynomials. Indeed if

P(x) = a

+ a

x + ···+ a

is a polynomial and f a function, then

(k)

(0) = f

(k)

(0) ⇐⇒ a

(k)

(0)

If this holds for all k ≤ n, then P = p

is the n

Taylor polynomial of f ! With a little modiﬁcation,

we’ve proved the following:

Theorem 3.33. 1. If f (x) =

∑

(x − c)

is a power series deﬁned on a neighborhood of c, then

f (x) = f (x): the function is its own Taylor series!

2. The n

Taylor polynomial of f centered at x = c is the unique polynomial p

of degree ≤ n

whose value and ﬁrst n derivatives agree with those of f at x = c: that is

∀k ≤ n, p

(k)

(c)

This answers our ﬁrst motivating question: a function can equal at most one power series with a

given center. The second question requires a careful study of the remainder: we’ll do this shortly.

Examples 3.34 (Common Maclaurin Series). These should be familiar from elementary calculus.

Each function equals the given series form our previous discussions of power series: by the Theorem,

each series is immediately the Maclaurin series of the given function.

∞

∑

n=0

x ∈ R

1 − x

∞

∑

n=0

x ∈ (−1, 1)

sin x =

∞

∑

n=0

(−1)

(2n + 1)!

2n+1

x ∈ R ln( 1 + x) =

∞

∑

n=1

(−1)

n+1

x ∈ (−1, 1]

cos x =

∞

∑

n=0

(−1)

(2n)!

x ∈ R tan

−1

x =

∞

∑

n=0

(−1)

2n + 1

2n+1

x ∈ [−1, 1]

Examples 3.35 (Modifying Maclaurin Series). By substituting for x in a common series, we quickly

obtain new series.

1. Substitute x 7→ 7x in the Maclaurin series for sin x, to recover our earlier example

sin 7x =

∞

∑

n=0

(−1)

2n+1

(2n + 1)!

2n+1

, x ∈ R

Note how this requires almost no calculation: since the function equals a series, the Theorem

says we have the Maclaurin series for sin 7x!

2. Substitute x 7→ x

in the Maclaurin series for e

to obtain

= exp(x

) =

∞

∑

n=0

, x ∈ R

This would be disgusting to verify directly, given the difﬁculty of repeatedly differentiating e

3. We ﬁnd the Taylor series for f (x) =

5−x

centered at x = 2:

f (x) =

3 + 2 − x

3( 1 −

x−2

)

∞

∑

n=0



x −2



which is valid whenever −1 <

x−2

< 1 ⇐⇒ −1 < x < 5.

4. Fix c ∈ R and observe that, for all x ∈ R,

= e

c+x−c

= e

x−c

∞

∑

n=0

(x −c)

We conclude that the series is the Taylor series of e

centered at x = c. Of course this is easily

veriﬁed using the deﬁnition, since



x=c

= e

5. Combining the Theorem with the multiple-angle formula, we obtain the Taylor series for sin x

centered at x = c:

sin x = sin(c + x −c) = sin c cos(x −c) + cos c sin(x −c)

∞

∑

n=0

(−1)

sin c

(2n)!

(x −c)

∞

∑

n=0

(−1)

cos c

(2n + 1)!

(x −c)

2n+1

Deﬁnition 3.36. A function f is analytic on a its domain if every c ∈ dom f has a neighborhood on

which f (x) equals its Taylor series centered at c.

All the examples we’ve thus far seen are analytic on their domains; indeed the last two of Exam-

ples 3.35 prove this for the exponential and sine functions. Every analytic function is automatically

smooth (inﬁnitely differentiable), however the converse is false (Exercise 10). Analyticity is of greater

importance in complex analysis where (amazingly!) it is equivalent to complex-differentiability.

Accuracy of Taylor Approximations

Our ﬁnal goal is to estimate the accuracy of a Taylor polynomial as an approximation to its generating

function. Otherwise said, we want to estimate the size of the remainder R

(x) = f (x) − p

(x).

Theorem 3.37 (Taylor’s Theorem: Lagrange’s form). Suppose f is n + 1 times differentiable on an

open interval I containing c and let x ∈ I \ {c}. Then there exists ξ between c and x for which the

remainder centered at c satisﬁes

(x) =

(n+1)

(ξ)

(n + 1)!

(x −c)

n+1

Proof. For simplicity let c = 0. Fix x = 0, deﬁne a constant M

and a function g : I → R by

(x) =

(n + 1)!

n+1

and g(t) =

(n + 1)!

n+1

+ p

(t) − f (t) =

(n + 1)!

n+1

− R

(t)

Observe that

k ≤ n + 1 =⇒ g

(k)

(x) =

(n + 1 − k)!

n+1−k

+ p

(k)

(t) − f

(k)

(t) (∗)

=⇒ g

(k)

(0) = p

(k)

(0) − f

(k)

(0) = 0 if k ≤ n

where we invoked Theorem 3.33.

Now apply Rolle’s Theorem repeatedly (WLOG assume x > 0):

• ∃ξ

between 0 and x such that g

′

(ξ

) = 0.

• ∃ξ

between 0 and ξ

such that g

′′

(ξ

) = 0, etc.

• Iterate to obtain a sequence (ξ

) such that

0 < ξ

n+1

< ξ

< ··· < ξ

< x and g

(k)

(ξ

) = 0

Take ξ = ξ

n+1

and consider (∗): since deg p

≤ n, we see that

0 = g

(n+1)

(ξ) = M

− f

(n+1)

(ξ) =⇒ R

(x) = f (x) − p

(x) =

(n+1)

(ξ)

(n + 1)!

n+1

Corollary 3.38. Suppose f is smooth on an open interval I containing c and that all derivatives f

(n)

of all orders are bounded on I. Then f equals its Taylor series (centered at c) on I.

Proof. For simplicity, let c = 0. Suppose



(n+1)

(ξ)



≤ K for all ξ ∈ I. Choose any N >

and

observe that

n > N =⇒

(x)

≤

n+1

(n + 1)!

n+1

N!(N + 1) ···(n + 1)

≤





n+1−N

−−−→

n→∞

Examples 3.39. 1. The functions sine and cosine have derivatives bounded by 1 on R, and thus both

functions equal their Maclaurin series on R. This removes the need to have previously justiﬁed

these facts using the theory of differential equations.

2. The exponential function does not have bounded derivatives, however we can still apply Tay-

lor’s Theorem. For any ﬁxed x, ∃ξ between 0 and x such that

(x)



(n + 1)!

n+1



−−−→

n→∞

by the same argument in the Corollary. Thus e

equals its Maclaurin series on the real line (we

knew this already from Exercise 3.28.14).

3. Extending Example 3.32.3, we see that h(x) =

√

x has linear approximation (1

Taylor polyno-

mial) centered at c = 9

(x) = h(9) + h

′

(9)(x −9) = 3 +

(x −9)

This yields the simple approximation

√

10 ≈ p

(10) = 3 +

Taylor’s Theorem can be used to estimate its accuracy (remember to shift the center to 9!):

(10) =

′′

(ξ)

(10 −9)

= −

·2!

−3/2

= −

8ξ

3/2

for some ξ ∈ (9, 10)

Certainly ξ

−3/2

< 9

−3/2

, whence

−

216

< R

(10) < 0 =⇒

−

216

683

216

√

10 <

684

216

is therefore an overestimate for

√

10, but is accurate to within

216

< 0.005.

Alternative Versions of Taylor’s Theorem

There are two further common expressions for the remainder in Taylor’s Theorem. These are typ-

ically less easy to use than Lagrange’s form but can sometimes provide sharper estimates for the

remainder, particularly when x is far from the center of the series.

Corollary 3.40. Suppose f

(n+1)

is continuous on an open interval I containing c, let x ∈ I \ {c}, and

let R

(x) = f (x) − p

(x) be the remainder for the Taylor polynomial centered at c. Then:

1. (Integral Remainder) R

(x) =

(x −t)

(n+1)

(t) dt

2. (Cauchy’s Form) ∃ξ between c and x such that R

(x) =

(x −ξ)

(x −c) f

(n+1)

(ξ)

Using these expressions it is possible to explicitly prove Newton’s binomial series formula.

Corollary 3.41. If α ∈ R and

< 1, then

(1 + x)

= 1 +

∞

∑

n=1

α(α −1) ···(α −n + 1)

= 1 + αx +

α(α −1)

α(α −1)(α −2)

α(α −1)(α −2)(α −3)

+ ···

If α ∈ N

, this is the usual binomial theorem. Otherwise it is more interesting: for instance,

√

1 + x = (1 + x)

1/2

= 1 +

x −

−

128

+ ···

(1 + x)

= 1 −3x + 6x

−10x

+ 15x

−···

Of course this last could easily be obtained from

1+x

∑

(−1)

by differentiating twice!

Exercises 3.31. Key concepts: Taylor Series/Polynomials, Lagrange’s form for Remainder

1. Compute the Maclaurin series for cos x directly from the deﬁnition and use Taylor’s Theorem

to indicate why it converges to cos x for all x ∈ R.

2. Repeat the previous exercise for sinh x =

−e

−x

) and cosh x =

+ e

−x

3. Find the Maclaurin series for the function sin( 3x

). How do you know you are correct?

4. Find the Taylor series of f (x) = x

−3x

+ 2x −5 at x = 2 and show that T

f (x) = f (x).

5. Find a rational approximation to

√

9 using the ﬁrst Taylor polynomial for f (x) =

√

x. Now use

Taylor’s Theorem to estimate its accuracy.

6. If c = 1, use the fact that 1 − x = (1 −c)



1 −

x−c

1−c



to obtain the Taylor series of

1−x

centered at

c. Hence conclude that

1−x

is analytic on its domain R \{1}.

7. We prove that the Maclaurin series

∑

∞

n=1

(−1)

n+1

converges to ln(1 + x) whenever 0 < x ≤ 1.

(a) Explicitly compute

n+1

ln( 1 + x).

(b) Suppose 0 < x ≤ 1. Using Taylor’s Theorem, prove that lim

n→∞

(x) = 0.

(If −1 < x < 0, the argument is tougher, being similar to Exercise 11)

8. Why can’t we use Taylor’s Theorem to approximate the error in

1−x

= 1 + x + R

(x) when

x ≥ 1? Try it when x = 2, what happens? What about when x = −2?

9. Prove Taylor’s Theorem with integral remainder when c = 0 by using the following as an

induction step: for each n ∈ N, deﬁne

(x) =

(x −t)

(n+1)

(t) dt

and use integration by parts to prove that A

n+1

= A

−

n+1

(n+1)!

(n+1)

(0).

(The Cauchy form follows from the intermediate value theorem for integrals which we’ll see later)

10. Consider the function

f (x) =

(

−1/x

if x > 0

0 otherwise

(a) Prove by induction that there exists a degree 2n polynomial q

for which

(n)

(x) = q





−1/x

whenever x > 0

(b) Prove that f is inﬁnitely differentiable at x = 0 with f

(n)

(0) = 0 (use Exercise 3.30.3).

The Maclaurin series of f is identically zero! Moreover, f is smooth (inﬁnitely differentiable) on R but

non-analytic at zero since it does not equal its Taylor series on any open interval containing zero.

A modiﬁcation allows us to create bump functions, which ﬁnd wide use in analysis. If a < b, deﬁne

a,b

: x 7→ f (x − a) f (b −x)

This is smooth on R but non-zero only on the interval (a, b). A

further modiﬁcation involving two such functions g

a,b

creates

a smooth function on R which satisﬁes

a,b,ϵ

(x) =

(

0 if x ≤ a −ϵ or x ≥ b + ϵ

1 if a ≤ x ≤ b

This ‘switches on’ rapidly from 0 to 1 near a and switches off

similarly near b. By letting ϵ be small, we smoothly (but not

uniformly) approximate the indicator function on [a, b].

a,b,ǫ

(x)

aa − ǫ b b + ǫ

11. (Hard) We prove the binomial series formula (Corollary 3.41).

Let f (x) = (1 + x)

and g(x) = 1 +

∑

∞

n=1

where a

α(α−1)···(α−n+1)

. Our goal is to prove

that f = g on the interval (−1, 1).

(a) Check that f

(n)

(0) = n!a

so that g really is the Maclaurin series of f .

(b) i. Prove that the radius of convergence of g is 1.

ii. Prove that lim

n→∞

= 0 whenever

< 1.

iii. If

< 1 and ξ lies between 0 and x, prove that



x−ξ

1+ξ



≤

(Hint: write ξ = tx for some t ∈ (0, 1). . . )

(x)

< (n + 1)

n+1

(1 + ξ)

α−1

Hence conclude that g = f whenever

< 1.

(d) Here is an alternative argument for the full result:

i. Show that (n + 1)a

n+1

+ na

= αa

ii. Differentiate term-by-term to prove directly that g satisﬁes the differential equation

(1 + x)g

′

(x) = αg(x). Solve this to show that g = f whenever

< 1.

4 Integration

The theory of inﬁnite series addresses how to sum inﬁnitely many ﬁnite quantities. Integration, by

contrast, is the business of summing inﬁnitely many inﬁnitesimal quantities. Attempts to do both have

been part of mathematics for well over 2000 years, and the philosophical objections are just as old.

The development and increased application of calculus from the late 1600s onward spurred mathe-

maticians to put the theory on a ﬁrmer footing, though from Newton and Leibniz it took another 150

years before Bernhard Riemann (1856) provided a thorough development of the integral.

4.32 The Riemann Integral

The basic idea behind Riemann integration is to approximate area using a sequence of rectangles

whose width tends to zero. The following discussion illustrates the essential idea, which should be

familiar from elementary calculus.

Example 4.1. Suppose f (x) = x

is deﬁned on [0, 1].

For each n ∈ N, let ∆x =

and deﬁne x

= i∆x.

Above each subinterval [x

i−1

, x

], raise a rectangle of height

f (x

) = x

. The sum of the areas of these rectangles is the Rie-

mann sum with right-endpoints

∑

i=1

f (x

)∆x =

∑

i=1

n(n + 1)(2n + 1)

3n + 1

The Riemann sum with left-endpoints is deﬁned similarly:

∑

i=1

f (x

i−1

)∆x =

∑

i=1

(i −1)

−

3n −1

Since f is an increasing function, the area A under the curve

plainly satisﬁes

≤ A ≤ R

By the squeeze theorem, we conclude that A =

0 1

n =

0.365234

0 1

n =

0.302734

The example should feel convincing, though perhaps this is due to the simplicity of the function. To

apply this approach to more general functions, we need to be signiﬁcantly more rigorous.

Two of Zeno’s ancient paradoxes are relevant here: Achilles and the Tortoise concerns a convergent inﬁnite series,

while the Arrow Paradox toys with integration by questioning whether time can be viewed as a sum of instants. Perhaps

the most famous contemporary criticism comes from Bishop George Berkeley, who gave his name to the city and ﬁrst

UC campus: in 1734’s The Analyst, Berkeley savaged the foundations of calculus, describing the inﬁnitesimal increments

required in Newton’s theory of ﬂuxions (derivatives) as merely the “ghosts of departed quantities.”

Recall some basic identities:

∑

i=1

i =

n(n + 1),

∑

i=1

n(n + 1)(2n + 1),

∑

i=1

(n + 1)

Deﬁnition 4.2. A partition P = {x

, . . . , x

} of an interval [a, b] is a ﬁnite sequence for which

a = x

< x

< ··· < x

n−1

< x

= b

Choosing a sample point x

∗

in each subinterval [x

i−1

, x

] results in a tagged partition.

The mesh of the partition is mesh(P) := max ∆x

, the width ∆x

= x

−x

i−1

of the largest subinterval.

If f : [a, b] → R, the Riemann sum

∑

i=1

f (x

∗

) ∆x

evaluates the area of a family of n rectangles, as

pictured. The heights f (x

∗

) and thus areas can be negative or zero.

f (x)

∗

b = x

∗

a = x

b = x

In elementary calculus, one typically computes Riemann sums for equally-spaced partitions with left,

right or middle sample points. The ﬂexibility of tagged partitions makes applying Riemann’s deﬁni-

tion a challenge, so we instead consider two special families of rectangles.

Deﬁnition 4.3. Given a partition P of [a, b] and a bounded function f on [a, b], deﬁne

= sup

x∈[x

i−1

]

f (x) U( f , P) =

∑

i=1

∆x

= inf

x∈[x

i−1

]

f (x) L( f , P) =

∑

i=1

∆x

U( f , P) and L( f , P) are the upper and lower Darboux sums for

f with respect to P. The upper and lower Darboux integrals are

U( f ) = inf U( f , P) L( f ) = sup L( f , P)

where the supremum/inﬁmum are taken over all partitions.

Necessarily both integrals are ﬁnite.

We say that f is (Riemann) integrable on [a, b] if U( f ) = L( f ) .

We denote this value by

f or

f (x) dx

a x

Upper Darboux sum U( f , P)

a x

Lower Darboux sum L( f , P)

If the interval is understood or irrelevant, one often simply says that f is integrable and writes

f .

Intuitively, L( f , P) is the sum of the areas of rectangles built on P which just ﬁt under the graph of f .

It is also the inﬁmum of all Riemann sums on P. If f is discontinuous, then L( f , P) need not itself be

a Riemann sum, as there might not exist suitable sample points!

Examples 4.4. 1. We revisit Example 4.1 in this language.

Given a partition Q = {x

, . . . , x

} of [0, 1] and sample points x

∗

∈ [x

i−1

, x

], we compute the

Riemann sum for f (x) = x

∑

i=1

f (x

∗

) ∆x

∑

i=1

∗

)

− x

i−1

)

Since f is increasing, we have x

i−1

≤ (x

∗

)

≤ x

on each interval, whence

L( f , Q) =

∑

i=1

i−1

)

− x

i−1

) ≤

∑

i=1

∗

)

− x

i−1

) ≤

∑

i=1

)

− x

i−1

) = U( f , Q)

The Darboux sums are therefore the Riemann sums for left- and right-endpoints.

If we take Q

to be the partition with subintervals of equal width ∆x =

, then

U( f ) = inf

U( f , P) ≤ U( f , Q

) =

∑

i=1





∆x = R

is the right Riemann sum discussed originally. Similarly L( f ) ≥ L

. Since L

and R

both

converge to

as n → ∞, the squeeze theorem forces

≤ L( f ) ≤ U( f ) ≤ R

=⇒ L( f ) = U( f ) =

Otherwise said, f is integrable on [0, 1] with

dx =

2. Suppose f (x) = kx + c on [a, b], and that k > 0. Take

the evenly spaced partition P

where x

= a +

b−a

Since f is increasing, the upper Darboux sum is again

the Riemann sum with right-endpoints:

U( f , P

) = R

∑

i=1

f (x

)∆x

b − a

∑

i=1

k(b − a)

i + ak + c

ak + c

bk + c

U( f , P

)

b − a



k(b − a)

n(n + 1) + (ak + c)n



−−−→

n→∞

k(b − a)

+ (b −a)(ak + c) =

− a

) + c( b − a)

Similarly, the lower Darboux sum is the Riemann sum with left-endpoints:

L( f , P

) = L

b − a



k(b − a)

n(n −1) + (ak + c)n



−−−→

n→∞

− a

) + c( b − a)

As above, L

≤ L( f ) ≤ U( f ) ≤ R

and the squeeze theorem prove that f is integrable on [a, b]

with

f =

− a

) + c( b − a).

Now we have some examples, a few remarks are in order.

Riemann versus Darboux Deﬁnition 4.3 is really that of the Darboux integral. Here is Riemann’s deﬁ-

nition: f : [a, b] → R being integrable with integral

f means

∀ϵ > 0, ∃δ such that (∀P, x

∗

) mesh(P) < δ =⇒



∑

i=1

f (x

∗

)∆x

−



< ϵ

This is signiﬁcantly more difﬁcult to work with, though it can be shown to be equivalent to the

Darboux integral. We won’t pursue Riemann’s formulation further, except to observe that if

a function is integrable and mesh(P

) → 0, then

f = lim

n→∞

∑

i=1

f (x

∗

)∆x

: this allows us to

approximate integrals using any sample points we choose, hence why right-endpoints (x

∗

= x

)

are so common in Freshman calculus.

Monotone Functions Darboux sums are easy to compute for monotone functions. As in the examples,

if f is increasing, then each M

= f (x

), from which U( f , P) is the Riemann sum with right-

endpoints. Similarly, L( f , P) is the Riemann sum with left-endpoints.

Area If f is positive and continuous,

the Riemann integral

f serves as a deﬁnition for the area

under the curve y = f (x). This should make intuitive sense:

1. In the second example where we have a straight line, we obtain the same value for the

area by computing directly as the sum of a rectangle and a triangle!

2. For any partition P, the area under the curve should satisfy the inequalities

L( f , P) ≤ Area ≤ U( f , P)

But these are precisely the same inequalities satisﬁed by the integral itself!

L( f , P) ≤ L( f ) =

f = U( f ) ≤ U( f , P)

In the examples we exhibited a sequence of partitions (P

) where U( f , P

) and L( f , P

) converged to

the same limit. The remaining results in this section develop some basic properties of partitions and

make this limiting process rigorous.

Deﬁnition 4.5. If P ⊆ Q are both partitions of [a, b], we call Q a reﬁnement of P.

To reﬁne a partition, we simply throw some more points in!

Lemma 4.6. Suppose f : [a, b] → R is bounded.

1. If Q is a reﬁnement of P (on [a, b]), then

L( f , P) ≤ L( f , Q) ≤ U( f , Q) ≤ U( f , P)

2. For any partitions P, Q of [a, b], we have L( f , P) ≤ U( f , Q).

3. L( f ) ≤ U( f )

We’ll see in Theorem 4.17 that every continuous function is integrable.

Proof. 1. We prove inductively. Suppose ﬁrst that Q = P ∪{t} contains exactly one additional point

t ∈ (x

k−1

, x

). Write

= inf



f (x) : x ∈ [x

k−1

, t]



= inf



f (x) : x ∈ [t, x

]



m = inf



f (x) : x ∈ [x

k−1

, x

]



= min{m

, m

}

The Darboux sums L( f , P) and L( f , Q) are identical ex-

cept for the terms involving t. This results in extra area:

k−1

Extra area!

··· ···

L( f , Q) − L( f , P) = m

(t − x

k−1

) + m

−t) −m(x

− x

k−1

)

= (m

−m)(t − x

k−1

) + (m

−m)(x

−t) ≥ 0

More generally, since a reﬁnement Q is obtained by adding ﬁnitely many new points, induction

tells us that P ⊆ Q =⇒ L( f , P) ≤ L( f , Q) .

The argument for U( f , Q) ≤ U( f , P) is similar, and the middle inequality is trivial.

2. If P and Q are partitions, then P ∪ Q is a reﬁnement of both P and Q. By part 1,

L( f , P) ≤ L( f , P ∪ Q) ≤ U( f , P ∪Q) ≤ U( f , Q) (∗)

3. This is an exercise.

Theorem 4.7. Suppose f : [a, b] → R is bounded.

1. (Cauchy criterion) f is integrable ⇐⇒ ∀ϵ > 0, ∃P such that U( f , P) − L( f , P) < ϵ.

2. f is integrable ⇐⇒ ∃(P

)

n∈N

such that U( f , P

) − L( f , P

) → 0. In such a situation, both

sequences U( f , P

) and L( f , P

) converge to

f .

Part 1 is termed a ‘Cauchy’ criterion since it doesn’t mention the integral (limit).

Proof. We prove the Cauchy criterion, leaving part 2 as an exercise.

(⇒) Suppose f is integrable and that ϵ > 0 is given. Since inf U( f , Q) =

f = sup L( f , R), there

exist partitions Q, R such that

U( f , Q) <

f +

and L( f , R) >

f −

Let P = Q ∪ R and apply (∗): L( f , R) ≤ L( f , P) ≤ U( f , P) ≤ U( f , Q). But then

U( f , P) − L( f , P) ≤ U( f , Q) − L( f , R) = U( f , Q) −

f +

f − L( f , R) < ϵ

(⇐) Assume the right hand side. For every partition, L( f , P) ≤ L( f ) ≤ U( f ) ≤ U( f , P). Thus

0 ≤ U( f ) − L( f ) ≤ U( f , P) − L( f , P) < ϵ

Since this holds for all ϵ > 0, we see that U( f ) = L( f ): that is, f is integrable.

Examples 4.8. 1. Consider f (x) =

√

x on the interval [0, b]. We choose a sequence of partitions (P

)

that evaluate nicely when fed to this function:

= {x

, . . . , x

} where x





=⇒ ∆x

= x

− x

i−1



−(i −1)



(2i −1)b

Since f is increasing on [0, b], we see that

U( f , P

) =

∑

i=1

f (x

)∆x

∑

i=1

√

(2i −1)b

3/2

∑

i=1

−i

3/2



n(n + 1)(2n + 1) −

n(n + 1)



−−−→

n→∞

3/2

Similarly

L( f , P

) =

∑

i=1

f (x

i−1

)∆x

∑

i=1

(i −1)

√

(2i −1)b

3/2

∑

i=1

−3i + 1

3/2



n(n + 1)(2n + 1) −

n(n + 1) + n



−−−→

n→∞

3/2

Since the limits are equal, we conclude that f is integrable and

√

x dx =

3/2

0 b

√

Upper Sum U( f , P

)

0 b

√

Lower Sum L ( f , P

)

2. Here is the classic example of a non-integrable function. Let f : [a, b] → R to be the indicator

function of the irrational numbers,

f (x) =

(

1 if x ∈ Q

0 if x ∈ Q

Suppose P = {x

, . . . , x

} is any partition of [a, b] . Since any interval of positive length contains

both rational and irrational numbers, we see that

sup



f (x) : x ∈ [x

i−1

, x

]



= 1 =⇒ U( f , P) =

∑

i=1

− x

i−1

) = b −a =⇒ U( f ) = b − a

inf



f (x) : x ∈ [x

i−1

, x

]



= 0 =⇒ L( f , P) = 0 =⇒ L( f ) = 0

Since the upper and lower Darboux integrals differ, f is not (Riemann) integrable.

As any freshman calculus student can attest, if you can ﬁnd an anti-derivative, then the fundamen-

tal theorem of calculus (Section 4.34) makes evaluating integrals far easier. For instance, you are

probably desperate to write

3/2

= x

1/2

=⇒

√

x dx =

3/2



3/2

rather than computing Riemann/Darboux sums as in the previous example! However, in most prac-

tical situations, no easy-to-compute anti-derivative exists; the best we can do is to approximate using

Riemann sums for progressively ﬁner partitions. Thankfully computers excel at such tedious work!

Exercises 4.32. Key concepts: Darboux sums/integrals, Partitions, sample points & reﬁnements,

Cauchy & sequential criteria for integrability

1. Use partitions to ﬁnd the upper and lower Darboux integrals on the interval [0, b]. Hence prove

that the function is integrable and compute its integral.

(a) f (x) = x

(b) g(x) =

√

2. Repeat question 1 for the following two functions. You cannot simply compute Riemann sums

for left and right endpoints and take limits: why not?

(a) h(x) = x(2 − x) on [0, 2]

(Hint: choose a partition with 2n subintervals such that x

= 1 and observe that h(2 −x) = h(x))

(b) On the interval [0, 3], let k(x) =

(

2x if x ≤ 1

5 − x if x > 1

(Hint: this time try a partition with 3n subintervals)

3. Let f (x) = x for rational x and f (x) = 0 for irrational x. Calculate the upper and lower

Darboux integrals for f on the interval [0, b]. Is f integrable on [0, b]?

4. Prove part 3 of Lemma 4.6: L( f ) ≤ U( f ).

5. Prove part 2 of Theorem 4.7.

f is integrable ⇐⇒ ∃(P

)

n∈N

such that lim

n→∞



U( f , P

) − L( f , P

)



= 0

Moreover, prove that both U( f , P

) and L( f , P

) converge to

f .

6. (a) Reread Deﬁnition 4.3. What happens if we allow f : [a, b] → R to be unbounded?

(b) (Hard) Read “Riemann versus Darboux” on page 73. Explain why being Riemann integrable

also forces f to be bounded.

on P.

7. (If you like coding) Write a short program to estimate

f (x) dx using Riemann sums. This

can be very simple (equal partitions with right endpoints), or more complex (random partition

and sample points given a mesh). Apply your program to estimate

sin(x

−

√

) dx.

4.33 Properties of the Riemann Integral

The rough take-away of this long section is that everything you think is integrable probably is! Ex-

amples will be few, since we have not established many explicit values for integrals.

Theorem 4.9 (Linearity). If f , g are integrable and k, l are constant, then k f + lg is integrable and

k f + lg = k

f + l

Example 4.10. Thanks to examples in the previous section, we can now calculate, e.g.,

−3

√

x dx = 5 ·

·2

−3 ·

·2

3/2

= 20 −4

√

Proof. Suppose ϵ > 0 is given. By the Cauchy criterion (Theorem 4.7, part 1), there exist partitions

R, S such that

U( f , R) − L( f , R) <

and U(g, S) − L(g, S) <

If P = R ∪S, then both inequalities are satisﬁed by P (Lemma 4.6). On each subinterval,

inf f (x) + inf g(x) ≤ inf



f (x) + g(x)



and sup



f (x) + g(x)



≤ sup f (x) + sup g(x)

since the individual suprema/inﬁma could be ‘evaluated’ at different places. Thus

L( f , P) + L(g, P) ≤ L( f + g, P) ≤ U( f + g, P) ≤ U( f , P) + U(g, P)

whence U( f + g, P) − L( f + g, P) < ϵ and f + g is integrable. Moreover,

( f + g) −

f −

g ≤



U( f , P) −





U(g, P) −



< ϵ

Using lower Darboux integrals similarly obtains the other half of the inequality

−ϵ <

( f + g) −

f −

g < ϵ

Since this holds for all ϵ > 0, we conclude that

( f + g) =

f +

That k f is integrable with

k f = k

f is an exercise. Put these together for the result.

Corollary 4.11 (Changing endvalues). Suppose f is integrable on [a, b] and g : [a, b] → R satisﬁes

f (x) = g(x) on (a, b). Then g is also integrable on [a, b] and

g =

f .

Deﬁnition 4.12 (Integration on an open interval). A bounded function g : (a, b) → R is integrable if

it has an integrable extension f : [a, b] → R where f (x) = g(x) on (a, b). In such a case, we deﬁne

g :=

f .

The Corollary (its proof is an exercise) shows that the choice of extension is irrelevant.

Theorem 4.13 (Basic integral comparisons). Suppose f and g are integrable on [a, b]. Then:

1. f (x) ≤ g(x) =⇒

f ≤

2. m ≤ f (x) ≤ M =⇒ m(b − a) ≤

f ≤ M(b −a)

3. f g is integrable.

is integrable and



≤

5. max( f , g) and min( f , g) are both integrable.

Part 3 is not integration by parts since it doesn’t tell us how

f g relates to

f and

Proof. 1. Since g − f is positive and integrable, L(g − f , P) ≥ 0 for all partitions P. But then

0 ≤ inf L(g − f , P) = L(g − f ) =

g − f =

g −

2. Apply part 1 twice.

3. This is an exercise.

4. The integrability is an exercise. For the comparison, apply part 1 to −

≤ f ≤

5. Use max( f , g) =

( f + g) +

f − g

, etc., together with the previous parts.

Theorem 4.14 (Domain splitting). Suppose f : [a, b] → R and

let c ∈ (a, b). If f is integrable on both [a, c] and [c, b], then it is

integrable on [a, b] and

f =

f +

f (x)

a c b x

In light of this result, it is conventional to allow integral limits to be reversed: if a < b, then

f := −

f is consistent with

f = 0

Proof. Let ϵ > 0 be given, then ∃R, S partitions of [a, c], [c, b] such that

U( f , R) − L( f , R) <

, U( f , S) − L( f , S) <

Choose P = R ∪S to partition [a, b], then

U( f , P) − L( f , P) = U( f , R) + U( f , S) − L( f , R) − L( f , S) < ϵ

Moreover

f (x)

a c b x

}| {

f −

f ≤ U( f , P) − L( f , R) − L( f , S) = U( f , P) − L( f , P) < ϵ

Showing that this expression is greater than −ϵ is similar.

Example 4.15. If f (x) =

√

x on [0, 1] and f (x) = 1 on [1, 2], then

f =

√

x dx +

1 dx =

+ 1 =

Monotonic & Continuous Functions We establish the integrability of two large classes of functions.

Deﬁnition 4.16. A function f : [a, b] → R is:

Monotonic if it is either increasing (x < y =⇒ f (x) ≤ f (y)) or decreasing.

Piecewise monotonic if there is a partition P = {x

, . . . , x

} (ﬁnite!) of [a, b] such that f is monotonic

on each open subinterval (x

k−1

, x

Piecewise continuous if there is a partition such that f is uniformly continuous on each (x

k−1

, x

Theorem 4.17. If f is monotonic or continuous on [a, b], then it is integrable.

Examples 4.18. 1. Since sine is continuous, we can approximate via a sequence of Riemann sums

sin x dx = lim

n→∞

∑

i=1

sin

πi

Evaluating this limit is another matter entirely, one best handled in the next section...

2. Similarly, e

√

is integrable and therefore may be approximated via Riemann sums:

√

dx = lim

n→∞

∑

i=1

exp

= lim

n→∞

∑

j=1

2j −1

exp

Both sums use right endpoints: the ﬁrst has equal subintervals, while the second is analogous

to Example 4.8.1. These limits would typically be estimated using a computer.

Proof. Since [a, b] is closed and bounded, a continuous function f is uniformly so. Let ϵ > 0 be given:

∃δ > 0 such that ∀x, y ∈ [a, b],

x −y

< δ =⇒

f (x) − f (y)

b − a

Let P be a partition with mesh P < δ. Since f attains its bounds on each [x

i−1

, x

∃x

∗

, y

∗

∈ [x

i−1

, x

] such that M

−m

= f (x

∗

) − f (y

∗

) <

b − a

from which

U( f , P) − L( f , P) <

∑

i=1

b − a

− x

i−1

) = ϵ

The monotonicity argument is an exercise.

Combining the proof with Deﬁnition 4.12: every uniformly continuous f : (a, b) → R is integrable.

Corollary 4.19. Piecewise continuous and bounded piecewise monotonic functions are integrable.

Proof. If f is piecewise continuous, then the restriction of f to (x

k−1

, x

) has a continuous extension

: [x

k−1

, x

] → R; this is integrable by Theorem 4.17. By Corollary 4.11, f is integrable on [x

k−1

, x

]

with

k−1

f =

k−1

. Theorem 4.14 (n −1 times!) ﬁnishes things off:

f =

∑

k=1

k−1

The argument for piecewise monotonicity is similar.

Example 4.20. The ‘fractional part’ function f (x) = x − ⌊x⌋

is both piecewise continuous and piecewise monotone on any

bounded interval. It is therefore integrable on any such interval.

0 1 2 3 4 5

For a ﬁnal corollary, here is one more incarnation of the intermediate value theorem.

Corollary 4.21 (IVT for integrals). If f is continuous on [a, b], then ∃ξ ∈ (a, b) for which

f (ξ) =

b − a

Proof. Since f is continuous, it is integrable on [a, b]. By the extreme value theorem it is also bounded

and attains its bounds: ∃p, q ∈ [a, b] such that

f (p) := inf

x∈[a,b]

f (x), f (q) = sup

x∈[a,b]

f (x)

Applying Theorem 4.13, part 2, with m = f (p) and

M = f (q), we see that

(b − a) f (p) ≤

f ≤ (b −a) f (q)

ξa bp q

Divide by b − a and apply the usual intermediate value theorem for f to see that the required ξ exists

between p and q.

In the picture, when f is positive and continuous, the grey area equals that under the curve; imagine

levelling off the blue hill with a bulldozer. . . The notation f

b−a

f indicates the average value

of f on [a, b]: to see why this interpretation is sensible, take a sequence of Riemann sums on equally-

spaced partitions P

(∆x =

b−a

) to see that

b − a

f = lim

n→∞

∑

i=1

f (x

∗

)

∆x

b − a

= lim

n→∞

f (x

∗

) + ··· + f (x

∗

)

is the limit of a sequence of averages of equally-spaced samples f (x

∗

What can/cannot be integrated?

We now know a great many examples of integrable functions:

• Piecewise continuous & monotonic functions are integrable.

• Linear combinations, products, absolute values, maximums and minimums of (already) inte-

grable functions.

By contrast, we’ve only seen one non-integrable function (Example 4.8.2). After so many positive

integrability conditions, it is reasonable to ask precisely which functions are Riemann integrable.

Here is the answer, though it is quite tricky to understand.

Theorem 4.22 (Lebesgue). Suppose f : [a, b] → R is bounded. Then

f is Riemann integrable ⇐⇒ it is continuous except on a set of measure zero

ıvely, the measure of a set is the sum of the lengths of its maximal subintervals, though unfortu-

nately this doesn’t make for a very useful deﬁnition.

Any countable subset has measure zero, so

Lebesgue’s result is almost as if we can extend Corollary 4.19 to allow for inﬁnite sums. For instance,

Exercise 1.17.8 describes a function which is continuous only on the irrationals: it is thus Riemann

integrable (indeed

f = 0 for any a < b). There are also uncountable sets with measure zero such

as Cantor’s middle-third set C: the function

f (x) =

(

1 if x ∈ C

0 otherwise

is continuous except on C and therefore Riemann integrable; again

f (x) dx = 0.

Exercises 4.33. Key concepts: Linear combinations, products, etc., of integrable functions are integrable,

Continuous and monotone functions are integrable, Integrability on open intervals

1. Explain why

2π

sin

) dx ≤

2. If f is integrable on [a, b] prove that it is integrable on any interval [c, d] ⊆ [a, b].

3. We complete the proof of Theorem 4.9 (linearity of integration).

(a) Suppose k > 0, let A ⊆ R and deﬁne kA := {kx : x ∈ A}. Prove that sup kA = k sup A

and inf kA = k inf A.

(b) If k > 0 prove that k f is integrable on any interval and that

k f = k

f .

Formally, the length of an open interval (a, b) is b − a and a set A ⊆ R has measure zero if

∀ϵ > 0, ∃ open intervals I

such that A ⊆

∞

[

n=1

and

∞

∑

i=1

length(I

) < ϵ

More generally, the Lebesgue measure of a set (subject to a technical condition) is the inﬁmum of the sum of the lengths of

any countable collection of open covering intervals. Measure theory is properly a matter for graduate study. Surprisingly,

there exist sets with positive measure that contain no subintervals, and even sets which are non-measurable!

4. Give an example of an integrable but discontinuous function on a closed bounded interval [a, b]

for which the conclusion of the Intermediate Value Theorem for Integrals is false.

5. Use Darboux sums to compute the value of the integral

15/2

1/2

x −⌊x⌋dx (Example 4.20).

6. We prove and extend Corollary 4.11. Suppose f is integrable on [a, b].

(a) If g : [a, b] → R satisﬁes f (x) = g(x) for all x ∈ (a, b), prove that g is integrable and

g =

f .

(Hint: consider h = f − g and show that

h = 0)

(b) Now suppose g : [a, b] → R satisﬁes f (x) = g(x) for all x ∈ [a, b] except at ﬁnitely many

points. Prove that g is integrable and

g =

f .

7. Show that an increasing function on [a, b] is integrable and thus complete Theorem 4.17.

(Hint: Choose a partition with mesh P <

f (b)−f (a)

)

8. Suppose f and g are integrable on [a, b].

(a) Deﬁne h(x) =



f (x)



. We know:

• f is bounded: ∃K such that

f (x)

≤ K on [a, b].

• Given ϵ > 0, ∃P such that U( f , P) − L( f , P) <

. For each subinterval [x

i−1

, x

], let

= sup f (x), m

= inf f (x), M

= sup h(x), m

= inf h(x)

Prove that M

−m

≤ 2(M

−m

)K. Hence conclude that h is integrable.

(b) Prove that f g is integrable.

(Hint: f g =

( f + g)

−

( f − g)

)

, P) − L(

, P) ≤ U( f , P) − L( f , P) for any partition P. Hence conclude

that

is integrable.

(One can extend these arguments to show that if j is continuous, then j ◦ f is integrable. Parts (a) and

and j(x) =

9. (Hard) Let f (x) =











x if x = 0 and sin

> 0

−x if x = 0 and sin

< 0

0 if x = 0

(a) Show that f is not piecewise continuous on [0, 1].

(b) Show that f is not piecewise monotonic on [0, 1].

(Hint: given ϵ, hunt for a suitable partition to make U( f , P) − L( f , P) < ϵ by considering [0, x

]

differently to the other subintervals)

(d) Make a similar argument which proves that g = sin

is integrable on ( 0, 1].

(Hint: Show that g has an integrable extension on [0, 1])

4.34 The Fundamental Theorem of Calculus

The key result linking integration and differentiation is usually presented in two parts. While there

are signiﬁcant subtleties, the rough statements are as follows (we follow the traditional numbering):

Part I Differentiation reverses integration:

f (t) dt = f (x)

Part II Integration reverses differentiation:

′

(x) dx = F( b) − F(a)

These facts seemed intuitively obvious to early practitioners of calcu-

lus. Given a continuous positive function f :

• Let F(x) denote the area under y = f (x) between 0 and x.

• A small increase ∆x results in the area increasing by ∆F.

• ∆F ≈ f (x)∆x is approximately the area of a rectangle, whence

∆F

∆x

≈ f (x). This is part I.

• F(b) − F(a) ≈

∑

∆F

≈

∑

f (x

)∆x

. Since F

′

= f , this is part II.

∆F

∆x

f (x)

When Leibniz introduced the symbols

and d in the late 1600s, it was partly to reﬂect the fundamen-

tal theorem.

If you’re happy with non-rigorous notions of limit, rate of change, area, and (inﬁnite)

sums, the above is all you need!

Of course we are very much concerned with the details: What must we assume about f and F, and

how are these properties used in the proof?

Theorem 4.23 (FTC, part I). Suppose f is integrable on [a, b]. For any x ∈ [a, b], deﬁne

F(x) :=

f (t) dt

Then:

1. F is uniformly continuous on [a, b];

2. If f is continuous at c ∈ [a, b], then F is differentiable

at c with F

′

Compare this with the na

ıve version above where we assumed f was continuous. We now require

only the integrability of f , and its continuity at one point for the full result.

is a stylized S for sum, while d stands for difference. Given a sequence F = (F

, F

, . . . , F

), construct a new

sequence of differences

dF = (F

− F

, F

− F

, . . . , F

− F

n−1

)

which can then be summed:

dF = (F

− F

) + (F

− F

) + ···(F

− F

n−1

) = F

− F

(∗)

Viewing a function as an ‘inﬁnite sequence’ of values spaced along an interval, dF becomes a sequence of inﬁnitesimals and

(∗) is essentially the fundamental theorem:

dF = F(b) − F(a). It is the concept of function that is suspect here, not the

essential relationship between sums and differences.

Strictly: if c = a, then F is right-differentiable, etc.

Examples 4.24. Examples in every elementary calculus course.

1. Since f (x) = sin

−7) is continuous on any bounded interval, we conclude that

sin

−7) dt = sin

−7)

If one follows Theorem 4.14 and its conventions, then this is valid for all x ∈ R.

2. The chain rule permits more complicated examples. For instance: f (t) = sin

√

t is continuous

on its domain [0, ∞) and y(x) = x

+ 3 has range [3, ∞) ⊆ dom( f ), whence

sin

√

t dt =

sin

√

t dt = 2x sin

+ 3

3. For a ﬁnal positive example, we consider when

sin x

tan(t

) dt = e

tan(e

) −cos x tan(sin

Makes sense. To evaluate this, ﬁrst choose any constant a and write

sin x

−

sin x

before differentiating. This is valid provided sin x, e

and a all lie in the same subinterval of

dom tan(t

) = R \ {±

, ±

3π

, ±

5π

, . . .}

Since

sin x

≤ 1 <

, this requires



⇐⇒ x <

Choosing a = 1 would certainly sufﬁce.

4. Now consider why the theorem requires continuity. The piecewise

continuous function

f : [0, 2] → R : x 7→

(

2x if x ≤ 1

if x > 1

has a jump discontinuity at x = 1. We can still compute

F(x) =

(

2t dt = x

if x ≤ 1

2t dt +

dt =

(x + 1) if x > 1

This is continuous, indeed uniformly so! However the discontinu-

ity of f results in F having a corner and thus being non-differentiable

at x = 1. Indeed F

′

(x) = f (x) whenever x = 1: that is, at all values

of x where f is continuous.

f (x)

0 1 2

F(x)

0 1 2

Proving FTC I Neither half of the theorem is particularly difﬁcult once you write down what you

know and what you need to prove. Here are the key ingredients:

1. Uniform continuity for F means we must control the size of

F(y) − F(x)



f (t) dt −

f (t) dt



f (t) dt



≤

f (t)

But the boundedness of f allows us to control this last integral. . .

2. F

′

x→c

F(x)−F(c)

x−c

= f (c), which means controlling the size of



F(x) − F(c)

x −c

− f (c)



x −c

f (t) dt − f (c)



The trick here will is to bring the constant f (c) inside the integral as

x−c

f (c) dt so that the

above becomes

x−c

f (t) − f (c)

dt. This may now be controlled via the continuity of f . . .

Proof. 1. Since f is integrable, it is bounded: ∃M > 0 such that

f (x)

≤ M for all x.

Let ϵ > 0 be given and deﬁne δ =

. Then, for any x, y ∈ [a, b],

0 < y − x < δ =⇒

F(y) − F(x)



f (t) dt



≤

f (t)

dt (Theorem 4.13, part 4)

≤ M(y −x) (Theorem 4.13, part 2)

< Mδ = ϵ

We conclude that F is uniformly continuous on [a, b] .

2. Let ϵ > 0 be given. Since f is continuous at c, ∃δ > 0 such that, for all t ∈ [a, b] ,

t − c

< δ =⇒

f (t) − f (c)

Now for all x ∈ [a, b] (except c),

0 <

x −c

< δ =⇒



F(x) − F(c)

x −c

− f (c)



x −c

f (t) − f (c) dt



(Theorem 4.9)

≤

x −c

f (t) − f (c)

dt (Theorem 4.13)

≤

x −c

< ϵ

Clearly lim

x→c

F(x)−F(c)

x−c

= f (c). Otherwise said, F is differentiable at c with F

′

The Fundamental Theorem, part II As with part I, the formulaic part of the result should be familiar,

though we are more interested in the assumptions and where they are needed.

Theorem 4.25 (FTC, part II). Suppose g is continuous on [a, b] , differentiable on (a, b), and moreover

that g

′

is integrable on (a, b) (recall Deﬁnition 4.12). Then,

′

= g(b) − g(a)

Part II is often expressed in terms of anti-derivatives: F being an anti-derivative of f if F

′

= f . Com-

bined with FTC, part I, we recover the familiar ‘+c’ result and a simpler version of the fundamental

theorem often seen in elementary calculus.

Corollary 4.26. Let f be continuous on [a, b].

• If F is an anti-derivative of f , then

f = F(b) − F(a).

• Every anti-derivative of f has the form F(x) =

f (t) dt + c for some constant c.

Examples 4.27. Again, basic examples should be familiar.

1. Plainly g(x) = x

+ 2x

3/2

is continuous on [1, 4] and differentiable on (1, 4) with derivative

′

(x) = 2x + 3

√

x; this last is continuous (and thus integrable) on (1, 4). We conclude that

2x + 3

√

x dx = x

+ 2x

3/2



= (16 + 16) − (1 + 2) = 29

2. If g(x) = sin(3x

), then g

′

(x) = 6x cos(3x

). Certainly g satisﬁes the hypotheses of the theorem

on any bounded interval [a, b] . We conclude

6x cos(3x

) dx = sin(3b

) −sin(3a

)

Moreover, every anti-derivative of f (x) = 6x cos(3x

) has the form F(x) = sin(3x

) + c.

3. Recall Example 4.24.4 where the discontinuity of f at x = 1 led to the non-differentiability of

F(x) =

f (t) dt. The function F therefore fails the hypotheses of FTC II on the interval [0, 2].

It almost, however, satisﬁes the conclusions of FTC II, though this is somewhat tautological

given the deﬁnition of F: except at x = 1, F is certainly an anti-derivative of f , and moreover

f (x) dx = F(2) − F(0).

In case you’re worried that this makes the theorem trivial, note that other anti-derivatives

F of

f exist (except at x = 1) which fail to satisfy the conclusion. For instance

F(x) =

(

if x < 1

x if x > 1

=⇒

F(2) −

F(0) = 1 =

f (x) dx

Proving FTC II Exercise 10 offers a relatively easy proof when g

′

= f is continuous. For the real

McCoy, we can only rely on the integrability of g

′

: the trick is to use the mean value theorem to write

g(b) − g(a) as a Riemann sum over a suitable partition.

Proof. Suppose ϵ > 0 is given. Since g

′

is integrable, we may choose some partition P satisfying

U(g

′

, P) − L(g

′

, P) < ϵ. Since g satisﬁes the mean value theorem on each subinterval,

∃ξ

∈ (x

i−1

, x

) such that g

′

(ξ

) =

g(x

) − g(x

i−1

)

− x

i−1

from which

g(b) − g(a) =

∑

i=1

g(x

) − g(x

i−1

) =

∑

i=1

′

(ξ

)(x

− x

i−1

)

This is a Riemann sum for g

′

associated to the partition P. Since the upper and lower Darboux sums

are the supremum and inﬁmum of these, we see that

L(g

′

, P) ≤ g(b) − g(a) ≤ U(g

′

, P)

However

′

satisﬁes the same inequality: L(g

′

, P) ≤

′

≤ U(g

′

, P). Since these inequalities

hold for all ϵ > 0, we conclude that

′

= g(b) − g(a).

While we certainly used the integrability of g

′

in the proof, it might seem strange that we assumed it

at all: shouldn’t every derivative be integrable? Perhaps surprisingly, the answer is no! If you want

a challenge, look up the Volterra function, which is differentiable everywhere but whose derivative is

non-integrable!

The Rules of Integration

If one wants to evaluate an integral, rather than merely show it exists, there are really only two options:

1. Evaluate Riemann sums and take limits. This is often difﬁcult if not impossible to do explicitly.

2. Use FTC II. The problem now becomes the ﬁnding of anti-derivatives, for which the core method

is essentially guess and differentiate. To obtain general rules, we can attempt to reverse the rules

of differentiation.

Integration by Parts Recall the product rule: the product g = uv of two differentiable functions is

differentiable with g

′

= u

′

v + uv

′

. Now apply Theorems 4.9, 4.13 and FTC II.

Corollary 4.28 (Integration by Parts). Suppose u, v are continuous on [a, b], differentiable on (a, b),

and that u

′

, v

′

are integrable on (a, b). Then

′

(x)v(x)dx = u(b)v(b) − u(a)v(a) −

u(x)v

′

(x)dx

This is signiﬁcantly less useful than the product rule since it merely transforms the integral of one

product into the integral of another.

Examples 4.29. With practice, there is no need to explicitly state u and v.

1. Let u(x) = x and v

′

(x) = cos x. Then u

′

(x) = 1 and v(x) = sin x. These certainly satisfy the

hypotheses. We conclude

π/2

x cos x dx =

[

x sin x

]

π/2

−

π/2

sin x dx =

sin

−0 −

[

−cos x

]

π/2

+ cos

−cos 0 =

−1

2. Let u(x) = ln x and v

′

(x) = 1. Then u

′

(x) =

and v(x) = x, whence

ln x dx =

[

x ln x

]

−

dx = e

ln e

−e ln e −

[

]

= 2e

−e −e

+ e = e

Change of Variables/Substitution We now turn our attention to the chain rule. If g(x) = F



u(x)



where F and u are differentiable, then g is differentiable with

′

(x) =

= F

′



u(x)



′

(x)

Now integrate both sides; the only issue is what assumptions are needed to invoke FTC II.

Theorem 4.30 (Substitution Rule). Suppose u : [a, b] → R and f : range(u) → R are continuous.

Suppose also that u is differentiable on (a, b) with integrable derivative u

′

. Then



u(x)



′

(x) dx =

u(b)

u(a)

f (u) du

This is the famous ‘u-sub’/change-of-variables formula from elementary calculus.

Proof. We leave as an exercise the veriﬁcation that both integrals exist. By the intermediate and

extreme value theorems, range(u) is a closed bounded interval. Assume range(u) has positive length

for otherwise both integrals are trivially zero.

Choose any c ∈ range(u) and deﬁne

F : range( u) → R by F(v) :=

f (t) dt

Since f is continuous, by FTC I says that F is differentiable with F

′

(u) = f ( u). But now



u(x)



′

(x) dx =





u(x)





dx (chain rule)

= F



u(b)



− F



u(a)



(FTC II)

u(b)

u(a)

f (u) du

Examples 4.31. Successfully applying the substitution rule can require signiﬁcant creativity.

1. To evaluate

√

2x sin x

dx, we consider the substitution u(x) = x

deﬁned on [0,

√

π].

Certainly u is continuous; moreover its derivative u

′

(x) = 2x is integrable on (0,

√

π). Finally

f (u) = sin u is continuous on range(u) = [0, π]. The hypotheses are satisﬁed, whence

√

2x sin x

dx =

√



u(x)



′

(x) dx =

u(π)

u(0)

f (u) du =

sin u du

= −cos u



= 2

2. For the following integral, a simple factorization suggests the substitution u(x) = x

− 2.

Plainly u : [

√

3] → [0, 1] and u

′

(x) = 2x is integrable. Moreover, f (u) =

is continuous

on range(u) = [0, 1]. We conclude

√

−4x

+ 5

dx =

√

−2)

+ 1

dx =

+ 1

du = arctan u



3. The hypotheses on u really are all that’s necessary. In particular, u need not be left-/right-

differentiable at the endpoints of [a, b]. For instance, with f (u) = u

and u(x) =

√

x on [0, 4],

we easily verify

√

x dx =

√

dx =



u(x)



′

(x) dx =

f (u) du =

du =

4. Sloppy ‘substitutions’ might lead to utter nonsense. For instance, u(x) = x

suggests

−1

dx =

−1

2x dx =

du =

(ln 4 −ln 1) = ln 2

This is total gibberish: the ﬁrst integral does not exist since

is undeﬁned at 0 ∈ (−1, 2).

Thankfully, the hypotheses of the substitution rule prevent this: f (u) =

is not continuous

on range(u) = [0, 4].

While you are very unlikely to make precisely this mistake, the risk is real in more complicated

or abstract situations. . .

Hence the old adage, “Differentiation is a science, whereas integration is an art.” To illustrate by example, consider

f (x) = tan(e

cos(3x

) + 4x

). The derivative is easily found using the product and chain rules:

d f

1 + (e

cos(3x

) + 4x

)



cos(3x

) −6xe

sin(3x

) + 12x



By contrast, if you want to ﬁnd an explicit anti-derivative of f (x), the integration analogues (parts/substitution) are essen-

tially useless. Similarly, the integral

tan(e

cos(3x

) + 4x

) dx

is likely impossible to evaluate explicitly and can only be approximated, say by using Riemann sums.

Exercises 4.34. Key concepts: Complete statements of FTC parts I & II, Integration by Parts/Substitution

1. Calculate the following limits:

(a) lim

x→0

dt (b) lim

h→0

3+h

2. Let f (t) =











0 if t < 0

t if 0 ≤ t ≤ 1

4 if t > 1

(a) Determine the function F(x) =

f (t) dt and sketch it. Where is F continuous?

(b) Where is F differentiable? Calculate F

′

at the points of differentiability.

3. Let f be continuous on R.

(a) Deﬁne F(x) =

x+1

x−1

f (t) dt. Carefully show that F is differentiable on R and compute F

′

(b) Repeat for G(x) =

sin x

f (t) dt.

4. Recall Examples 4.24.4 and 4.27.3. Describe all anti-derivatives F of f on [0, 1) ∪ (1, 2]. Which

satisfy

f (x) dx = F(2) − F(0)?

5. Suppose u, v satisfy the hypotheses of integration by parts. By FTC I,

′

(t)v(t) dt is an anti-

derivative of u

′

(x)v(x): what does integration by parts say is another?

6. Use a substitution to integrate

√

1 − x

7. Use integration by parts and the substitution rule to evaluate

arcsin x dx for any b < 1.

8. Use integration by parts to evaluate

x arctan x dx for any b > 0

9. If f and u satisfy the hypotheses of the substitution rule, explain why both ( f ◦ u)u

′

and f are

integrable on the required intervals.

10. We prove a simpler version of the fundamental theorem when f : [a, b] → R is continuous.

Part I Deﬁne F(x) =

f (t) dt. If c, x ∈ [a, b] where c = x, prove that

m ≤

F(x) − F(c)

x −c

≤ M

where m, M are the maximum and minimum values of f (t) on the closed interval with

endpoints c, x; why do m, M exist? Now deduce that F

′

Part II Now suppose F is any anti-derivative of f on [a, b]. Use part (a) and the mean value

theorem to prove that

f (t) dt = F(b) − F(a).

4.36 Improper Integrals

The Riemann integral has several limitations. Even allowing for functions to be integrable on open

intervals (Deﬁnition 4.12), the existence of

f (x) dx requires both:

• That (a, b) be a bounded interval.

• That f be bounded on (a, b).

Limits provide a natural way to extend the Riemann integral to unbounded intervals and functions.

Deﬁnition 4.32. Suppose f : [a, b) → R satisﬁes the following properties:

• f is integrable on every closed bounded subinterval [a, t] ⊆ [a, b).

• If b is ﬁnite, then f is unbounded at b (b can be ∞!)

The improper integral of f on [a, b) is

f (x) dx := lim

t→b

−

f (x) dx

This is convergent or divergent as is the limit.

If an integral is improper at its lower limit (f : (a, b] → R, etc.), then

f (x) dx := lim

s→a

f (x) dx.

If an integral is improper at both ends, choose any c ∈ (a, b) and deﬁne

f (x) dx = lim

s→a

f (x) dx + lim

t→b

−

f (x) dx

provided both one-sided improper integrals exist and the limit sum makes sense.

Theorem 4.14 says that the choice of c for a doubly-improper integral is irrelevant.

Many properties of the Riemann integral transfer naturally to improper integrals, though not every-

thing. . . For example, part 1 of Theorem 4.13 extends:

Theorem 4.33. If 0 ≤ f (x) ≤ g(x) on [a, b), then

f ≤

g whenever the integrals exist (standard

or improper). In particular:

•

f = ∞ =⇒

g = ∞

•

g convergent =⇒

f converges to some value ≤

We leave some of the detail to Exercise 7.

Examples 4.34. 1.

dx =

for any t > 0. Clearly

∞

dx = lim

t→∞

= ∞

More formally, the improper integral

∞

dx diverges to inﬁnity.

2. With f (x) = x

−4/3

deﬁned on [1, ∞),

∞

−4/3

dx = lim

t→∞

−4/3

dx = lim

t→∞

−3x

−1/3

= lim

t→∞

3 −3t

−1/3

= 3

3. Consider f (x) =

−x

on (−∞, ∞). On any bounded interval [ 0, t],

f (x) dx =

−x

dx =

−e

−x

= 1 − e

−t

−−→

t→∞

By symmetry,

∞

−∞

−x

dx = 1 + 1 = 2

This example arises naturally in probability: multiplying by

√

2π

computes the expectation of

when X is a standard normally-distributed random variable

) =

∞

−∞

√

2π

−x

dx =

4. Our knowledge of derivatives

sin

−1

x =

√

1−x

(or the substitution rule) allows us to evaluate

√

1 − x

dx = lim

t→1

−

√

1 − x

dx = lim

t→1

−

sin

−1

t =

By symmetry,

−1

√

1−x

dx = π. By comparison, we obtain bounds on another improper inte-

gral:

√

1 − x

≤

√

1 − x

=⇒

−1

√

1 − x

dx ≤

−1

√

1 − x

dx = π

5. Improper integrals need not exist. For instance,

lim

t→∞

sin x dx = lim

t→∞

1 −cos t

diverges by oscillation.

Exercises 4.36. Key concepts: Formal deﬁnition and careful calculation of Improper Integrals

1. Use your answers from Section 4.34 to decide whether the improper integrals

arcsin x dx

and

∞

x arctan x dx exist. If so, what are their values?

2. Let p be a positive constant. Prove:

dx =

(

1−p

if p < 1

∞ if p ≥ 1

∞

dx =

(

p−1

if p > 1

∞ if p ≤ 1

(The ﬁrst of these justiﬁes the convergence/divergence properties of p-series via the integral test)

3. Suppose f is integrable on [a, b]. Explain why

f (x) dx = lim

t→b

−

f (x) dx is still true, even

though the integral is not improper.

4. State a version of integration by parts modiﬁed for when

′

(x)v(x) dx is improper at b. Now

evaluate

∞

−4x

dx.

5. What is wrong with the following calculation?

∞

−∞

x dx = lim

t→∞



−t

= lim

t→∞

−t

) = lim

t→∞

0 = 0

6. Prove or disprove: if

f and

g are convergent improper integrals, so is

f g.

7. Prove part of Theorem 4.33. Suppose 0 ≤ f (x) ≤ g(x) for all x ∈ [a, b), and that

g is a

convergent improper integral. Prove that

f converges and that

f ≤

Extensions of the Riemann Integral (just for fun)

In the 1890s, Thomas Stieltjes

offered a generalization of the Riemann integral.

Deﬁnition 4.35. Let f : [a, b] → R be bounded and α : [a, b] → R monotonically increasing. Given a

partition P = {x

, . . . , x

} of [a, b], deﬁne the sequence of differences

∆α

= α(x

) − α(x

i−1

)

The upper/lower Darboux–Stieltjes sums/integrals are deﬁned analogously to the pure Riemann case:

U( f , P, α) =

∑

i=1

sup

i−1

]

f (x) ∆α

L( f , P, α) =

∑

i=1

inf

i−1

]

f (x) ∆α

U( f , α) = inf

U( f , P, α) L( f , α) = sup

L( f , P, α)

If U( f , α) = L( f , α), we say that f is Riemann–Stieltjes integrable of class R(α) and denote its value

f (x) dα.

The standard Riemann integral corresponds to α(x) = x. It is the ability to choose other functions α

that makes the Riemann–Stieltjes integral both powerful and applicable.

Standard Properties Most results in sections 4.32 and 4.33 hold with suitable modiﬁcations, as does

the discussion of improper integrals. For instance,

f ∈ R(α) ⇐⇒ ∃P such that U( f , P, α) − L( f , P, α) < ϵ

The result regarding the piecewise continuity of f is a notable exception: depending on α, a

piecewise continuous f might not lie in R(α).

Weighted integrals If α is differentiable, we obtain a standard Riemann integral

f (x) dα =

f (x)α

′

(x) dx

weighted so that f (x) contributes more when α is increasing rapidly.

Probability If α(a) = 0 and α(b) = 1, then α may be viewed as a probability distribution function and its

derivative α

′

as the corresponding probability density function. For example:

1. The uniform distribution on [a, b] has α =

b−a

(x − a) so that

f (x) dα =

b − a

f (x) dx

Since α

′

is constant, the integrals weigh all values of x uniformly.

2. The standard normal distribution has α(x) =

−∞

√

2π

−t

dt. The fact that α

′

√

2π

−x

is maximal when x = 0 reﬂects the fact that a normally distributed variable is clustered

near its mean.

In all cases,

f (x) dα = E( f (X)) computes an expectation (see, e.g., Example 4.34.3).

Stieltjes was Dutch; the pronunciation is roughly ‘steelchez.’

Non-differentiable or continuous α This provides major ﬂexibility! For example, if Q = {s

, . . . , s

}

partitions [a, b] , and (c

)

k=1

is a positive sequence, then

α(x) =











0 if x = a

∑

i=1

if x ∈ ( s

k−1

, s

]

deﬁnes an increasing step function, and the Riemann–Stieltjes integral a weighted sum

f (x) dα =

∑

i=1

f (s

)

Taking an inﬁnite increasing sequence (s

) ⊆ [a, b] results in an inﬁnite series, which helps

explain why so many results for series and integrals look similar!

This also touches on probability. For example, let p ∈ [0, 1], n ∈ N, and s

= k on the interval

[0, n]. If c

(

)

(1 − p)

n−k

, then

f (x) dα =

∑

k=0





(1 − p)

n−k

f (x) = E( f (X))

is the expectation of f (X) when X ∼ B(n, p) is binomially distributed.

Lebesgue Integration: Integrals and Convergence

Lebesgue’s extension essentially uses rectangles whose heights tend to zero: cutting up the area under

a curve using horizontal instead of vertical strips. One of its major purposes is to permit a more general

interchange of limits and integration in many cases of pointwise (non-uniform) convergence. To see

the problem, consider the sequence of piecewise continuous functions

: [0, 1] → R : x 7→

(

1 if x =

∈ Q with q ≤ n

0 otherwise

Each f

is Riemann integrable with

(x) dx = 0. However, the pointwise limit

f (x) =

(

1 if x ∈ Q

0 if x ∈ Q

is not Riemann integrable (compare Example 4.8.2). In the Lebesgue theory, the limit f turns out to

be integrable with integral 0, so that

lim

n→∞

(x) dx =

lim

n→∞

(x) dx

Recall (Theorem 2.19) that the interchange of limits and integrals would be automatic if the conver-

gence f

→ f were uniform: of course the convergence isn’t uniform here.

Like measure theory (recall Theorem 4.22), Lebesgue integration is a central topic in graduate analysis.