Math 140B - Notes

Neil Donaldson

January 22, 2025

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

( u

) = (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.)