Math 140B - Notes

Neil Donaldson

Fall 2022

1 Continuity

The primary goal of this course is to make elementary calculus rigorous. We begin with a review of

some basic concepts and conventions.

Sets & Functions We are concerned with functions f : U → V where both U, V are subsets of the

real numbers R:

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 intervals of positive length.

Codomain codom( f ) = V. We often take V = R by default.

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

Injectivity f is injective/one-to-one if f (x) = f (y) =⇒ x = y.

Surjectivity f is surjective/onto if f (U) = V.

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 domain/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 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.

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 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

→2

f (x

) = lim

→2



lim

→2



= 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

Section numbers are identical to those in the ofﬁcial textbook.

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 (Common Continuous Functions).

1. Suppose f and g are continuous at u, that h is continuous at f (u) and that k is constant. Then

the following are continuous at u (if deﬁned):

f + g, f − g, f g,

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

2. Algebraic

functions are continuous.

3. The common 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, ∞).

These claims are tedious to prove using elementary deﬁnitions. The ﬁrst two require many uses

of the limit laws, while the transcendental claim is easier to defer until we can deﬁne the common

functions using power series, after which continuity comeS for free.

Constructed using ﬁnitely many addition/subtraction, multiplication/division and n

root operations

Exercises 17 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. Suppose f and g are continuous at u. Prove the following using ϵ –δ arguments.

(a) f − g is continuous at u.

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

5. Contrary to our standing assumption, 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.

6. Refresh your prerequisites by giving formal proofs of the following:

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

) ⊆ U such that x

→ M.

(Remember that 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 exists a family of nested intervals I

⊇ I

⊇ ···

such that inﬁnitely many of the terms (x

) lie in each interval I

. Hence obtain a subsequence

) and prove that it is Cauchy.

)

7. (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.

This is a good moment to review the notion of a Cauchy sequence

∀ϵ > 0, ∃N such that m, n > N =⇒

− x

< ϵ

and the discussion of Cauchy completeness: (x

) ⊆ R is convergent if and only if it is Cauchy.

18 Properties of Continuous Functions

The goal of this section is to 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 17.6:

• (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 generalize to functions on arbitrary intervals. Our next result should be familiar from ele-

mentary calculus and is intuitively obvious from the na

ıve notion of continuity: graph such a func-

tion without taking your pen from 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 (see Exercise 18.2) 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 6, ∃(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 the proof, we can play a similar game with the sequence deﬁned by t

= min{b, ξ +

}; this

is left to Exercise 4.

Example 1.14. The intermediate value theorem is particularly useful for demonstrating the exis-

tence of solutions to equations. For example, we can use the following steps to show that the equation

= 1 has a solution.

• 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 <

We ﬁnish with a useful corollary.

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 an interval apart.

For example, f (x) =

√

−4 has domain (−∞, −2] ∪ [2, ∞) and range [0, ∞): both original intervals

get mapped to the same interval by f .

A more general statement from topology says that if f : U → V is continuous between topological

spaces and a, b lie in the same component of U, then f (a) and f (b) lie in the same component of f (U).

In single-variable real analysis each component is an interval.

Exercises 18 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.

19 Uniform Continuity

Recall the ϵ–δ deﬁnition of continuity: f : U → R is continuous at all points

y ∈ U, we require

∀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 on both y and ϵ. In the na

ıve sense of

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

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

Deﬁnition 1.16. 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) in both deﬁnitions to make clear that ϵ and

δ are independent of x and y. Note also that the deﬁnition is now symmetric in x and y.

Example 1.17. 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 above examples.

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.

Extendability In part 1 (if b = ∞), the domain of f may be extended: g : [a, b] → R : x 7→

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

that g(x) =

whenever x > 0.

If the gradient of a continuous function is bounded or if you can ‘ﬁll in the holes’ at the endpoints of

its domain, then the function is uniformly continuous. While the utility of uniform continuity is often

in proofs when the independence of ϵ and location are critical, it is often one of the above properties

that is being invoked. The remainder of this section involves making thse observations watertight.

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

Theorem 1.18. Let f : I → R be differentiable on an interval I. If the derivative f

′

is bounded on

the interior I

◦

, then f is uniformly continuous on I.

The proof depends on the mean value theorem, which we’ll prove later in the term.

Proof. Suppose

′

(x)

≤ M on I

◦

. Let ϵ > 0 be given, let δ =

and suppose x, y ∈ I with x > y.

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 1.18 isn’t a biconditional: for instance, Exercise 19.5 shows that f (x) =

√

x on [0, ∞) and

g(x) = x

1/3

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

We now discuss the idea of extendability and how uniform continuity relates to continuity on closed

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

Theorem 1.19. 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 to a partial converse of this.

Lemma 1.20. 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 it is

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

Theorem 1.21. 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.

Deﬁne a sequence

( u

) = (x

, y

, x

, y

, x

, y

, . . .)

Since (x

) and (y

) have the same limit a, we see that u

→ a. But then ( u

) is Cauchy;

by Lemma 1.20, ( 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.

(⇐) Certainly if g is well-deﬁned then it is continuous. By Theorem 1.19 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 1.22. 1. f : x 7→ x

is uniformly continuous on (−3, 10) since its derivative f

′

(x) = 2x

is bounded (

′

(x)

= 2

≤ 20) on its domain. It has the obvious continuous extension

g(x) = x

on [−3, 10].

2. Neither argument works for f (x) = x

on the domain (−3, ∞): both f

′

and the domain (−3, ∞)

are unbounded, so neither Theorem 1.18 nor 1.21 applies.

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.

3. f (x) = x sin

is continuous on the interval (0, ∞ ). Strictly, neither Theorem 1.18 nor 1.21

applies 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 1.18 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 1.21, f is uni-

formly continuous on ( 0, 1].

Putting this together, 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 19 1. Which of the following functions are uniformly continuous on the speciﬁed set?

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 of the following functions is uniformly continuous on the indicated set 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 that 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. (a) Let f (x) =

√

x with domain [0, ∞). Show that f

′

(x) is unbounded, but that f is still

uniformly continuous on [0, ∞).

(Hint: try δ = ϵ

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

√

y + ϵ)

. . . )

(b) Prove that g(x) = x

1/3

is uniformly continuous on R.

(Hint: try δ = (

)

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

20 Limits of Functions

You’ve likely seen many calculations of the following form in elementary calculus:

lim

x→3

−9

x −3

= lim

x→3

(x −3)(x + 3)

x −3

= lim

x→3

(x + 3) = 6

Our next goal is to make this notation precise and to tie it to our earlier notion of limit.

Deﬁnition 1.23. Suppose f : U → R, that S ⊆ U, and that a is the limit of a sequence

in S.

We write lim

x→a

f (x) = L and say that L is the limit of f (x) as x tends to a along S, provided

∀(x

) ⊆ S, lim x

= a =⇒ lim f (x

) = L

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

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

• The one-sided deﬁnitions apply when a = ±∞, though we omit the ± modiﬁers: for instance,

lim

x→∞

f (x) = L ⇐⇒ lim

x→∞

f (x) = L for some S = (c, ∞) ⊆ U

• The subtlety in the deﬁnition is that for lim

x→a

f (x) to be deﬁned, the domain U of f must contain

a punctured neighborhood S of a: i.e. a ∈ U

◦

. The one-sided limits similarly require a one-sided

punctured neighborhood. These conditions are always satisﬁed if U is a disjoint union of intervals

of positive length, in which case lim

x→a

(±)

f (x) = L if and only if

lim f (x

) = L, ∀(x

) ⊆ U \ {a} tending to a (from above/below)

In this situation, Deﬁnition 1.6 recovers the 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

(∗)

• By modifying the proof of Theorem 1.8 in the case that a, L ∈ R are ﬁnite, the above can be

written in ϵ-language. For example lim

x→a

f (x) = L means

∀ϵ > 0, ∃δ > 0 such that (∀x ∈ R) 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!

I.e. a ∈ S or perhaps a = ±∞ if S is unbounded.

Examples 1.24. 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 so often in the above examples, it is appropriate to update them

to this new context. We do so without proof.

Corollary 1.25 (Limit Laws). 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 1.26. 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. Being

able easily to equate these limits is one of the advantages of the ‘∃S’ formulation of Deﬁnition 1.23. Think about why; what

is a suitable set S in this context?

Classiﬁcation of Discontinuities

We ﬁnish this section by considering the ways in which a function can fail to be continuous.

Deﬁnition 1.27. Suppose that a function is continuous on an interval except at ﬁnitely many values:

we call these isolated discontinuities.

Examples 1.28. 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 1.29. Let f : U → R and suppose a ∈ U

◦

is an interior point. 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 1.30. 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 20 1. For the function f (x) =

, determine the limits 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 evaluate? Prove your assertions in each case.

(a) lim

x→∞

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

x→∞

( f − g)(x)

x→∞

( f g)(x) (d) lim

x→∞

( f /g)(x)

5. Complete the proof of Theorem 1.29 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. Let a ∈ U

◦

. Verify the claim following Deﬁnition 1.23: 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 17.5, 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 1.23.)