Limit of a Function - Kennesaw State Universityksuweb.kennesaw.edu/~plaval/math4381/func_lim.pdf · Limit of a Function 5.1 De–nitions and Examples 5.1.1 Overview In this chapter,

Chapter 5

Limit of a Function

5.1 Definitions and Examples

5.1.1 Overview

In this chapter, we consider a function f : D → R where D ⊆ R. We alsoconsider a real number a which is a limit point of D. We will investigate thefollowing limits: lim

x→±∞f (x) and lim

x→af (x). That is, we are studying how f

behaves near a point a or near infinity. The questions we are trying to answerare:

1. As x is getting closer and closer to a, how is y = f (x) behaving? Is itgetting closer to a number as well? Is it getting arbitrary large (in absolutevalue)? Is it not following any pattern?

2. Same question if x is approaching ∞ or −∞.

In more general terms, we are asking the question: if x is following a certainpattern, is f (x) also following a pattern and if yes, which pattern?In order to be able to evaluate lim

x→af (x), f must be defined in a deleted neigh-

borhood of a that is f must be defined in an interval of the form (a− h, a+ h)for some positive number h, except maybe at x = a.When we say lim

x→af (x) = L, we mean that f (x) can be made as close

as we want from L simply by taking x close enough to a. Or, in terms ofneighborhoods, we have the following general definition for a limit.

Definition 5.1.1 We say that limx→a

f (x) = L or that f (x) → L as x → a if

for every neighborhood V of L, one can find a deleted neighborhood U of a suchthat x ∈ U =⇒ f (x) ∈ V .

This definition can be adapted to limits at a finite point or at infinity aswell as when the limit is finite or infinite. There are three possibilities for x, we

163

164 CHAPTER 5. LIMIT OF A FUNCTION

can have x → a, x → −∞ and x → ∞. For each case, we can have f (x) → L,f (x) → −∞ and f (x) → ∞. Hence, we have a total of nine definitions. Theycan all be derived from the above definition simply by remembering that aneighborhood of a finite point a is an interval of the form (a− δ, a+ δ) and aneighborhood of infinity is an interval of the form (w,∞) for some w ∈ R.

5.1.2 Limit at a finite point


f (x) = L or that f (x) → L as x → a if

∀ε > 0, ∃δ > 0 : 0 < |x− a| < δ =⇒ |f (x)− L| < ε.

|x− a| represents how far x is from a. The above statement says that f (x)can be made arbitrarily close to L simply by taking x close enough to a.

Example 5.1.3 Prove that limx→2

(x+ 5) = 7.

Given ε > 0, we must prove that there exists δ > 0 such that 0 < |x− 2| < δ =⇒|x+ 5− 7| < ε. Let ε > 0 be given.

|x+ 5− 7| < ε⇐⇒ |x− 2| < ε

Thus we see that δ = ε will work. Indeed, given ε > 0, 0 < |x− 2| < δ =⇒|x+ 5− 7| < ε.

Remark 5.1.4 Of course, this was a very easy example to illustrate how thiskind of problem is addressed. In general, it will take more work to find δ givenε > 0. Many of the techniques used for sequences will also be used here. Somewill be illustrated below when we look at more challenging examples.

Remark 5.1.5 Saying that 0 < |x− a| < δ is the same as saying that x ∈(a− δ, a+ δ) and x 6= a. Similarly, saying that |f (x)− L| < ε is the same assaying that f (x) ∈ (L− ε, L+ ε).

Remark 5.1.6 In order to understand how to adapt this definition to casesinvolving ∞ such as in the case when L = ∞ or also when x → ∞, it isimportant to understand the notion of neighborhood of ∞. A neighborhood of∞ is an interval of the form (w,∞). Similarly, an neighborhood of −∞ is aninterval of the form (−∞, w).


f (x) = ∞ or that f (x) → ∞ as x → a if

∀M > 0, ∃δ > 0 : 0 < |x− a| < δ =⇒ f (x) > M .

Example 5.1.8 Prove that limx→1

1

(x− 1)2 =∞.

Given M > 0, we must prove that there exists δ > 0 such that 0 < |x− 1| <

5.1. DEFINITIONS AND EXAMPLES 165

δ =⇒ 1

(x− 1)2 > M .

1

(x− 1)2 > M ⇐⇒ (x− 1)

2<

1

M

⇐⇒ −1√M

< x− 1 <1√M

⇐⇒ |x− 1| < 1√M

Thus, given M > 0, δ =1√M

will work.


f (x) = −∞ or that f (x)→ −∞ as x→ a

if ∀M < 0, ∃δ > 0 : 0 < |x− a| < δ =⇒ f (x) < M .

Remark 5.1.10 If we find a δ which works, then every δ′ < δ will also work.Therefore, it is always possible to impose certain conditions on δ such as sayingthat we are looking for δ less than a certain number h, thus restricting our searchto an interval of the form (a− h, a+ h). In this interval, if we call δ′ the valuewe found, then δ = min

(h, δ′

).

Remark 5.1.11 In the last two definition, the vertical line x = a is a verticalasymptote for the graph of y = f (x).

Remark 5.1.12 In the definition of a limit, it is implied that a is a limit pointof D (f) that is for every δ > 0 the interval (a− δ, a+ δ) contains points ofD (f) other than a. If this is not the case, then for δ small enough, there maynot exist any x satisfying 0 < |x− a| < δ. In this case, the concept of a limithas no meaning.

Remark 5.1.13 In the definition of a limit, a does not have to be in the domainof f . It only needs to be a limit point of D (f).

Remark 5.1.14 In the definition of a limit, δ depends obviously on ε. It mayalso depend on the point a as illustrated by example 5.1.38.

Remark 5.1.15 When we say that the limit of a function exists, we mean thatit exists and is finite. When the limit is infinite, it does not exist in the sensethat it is not a number. However, we know what the function is doing, it isapproaching ±∞.Remark 5.1.16 There are several situations under which a limit will fail toexist.

1. The function may oscillate boundedly like in f (x) = sin1

xas x→ 0.

2. The function may oscillate unboundedly like x sinx as x→∞.

3. The function may grow without bounds like1

x2as x→ 0.

4. There may be a "break" in the graph.


Limit at infinity

In order to be able to evaluate limx→∞

f (x), f must be defined for large x. In

other words, we must have D (f)∩ (w,∞) 6= ∅ for every w ∈ R. In the case wewant to evaluate lim

x→−∞f (x), then we must have D (f)∩ (−∞, w) 6= ∅ for every

w ∈ R. We then have the following definitions (some of the definitions will beaccompanied with easy examples to illustrate the concept being defined):

Definition 5.1.17 We say that limx→∞

f (x) = L or that f (x)→ L as x→∞ if

∀ε > 0, ∃w > 0 : x ∈ (w,∞) ∩D (f) =⇒ |f (x)− L| < ε

|f (x)− L| represents the distance between f (x) and L. The above state-ment simply says that f (x) can be made as close as one wants to L, simply bytaking x large enough. Graphically, this simply says that the line y = L is ahorizontal asymptote for the graph of y = f (x).To prove that a number f (x) approaches L as x→∞, given ε > 0, one has

to prove that w > 0 can be found so that x ∈ (w,∞) ∩D (f) =⇒ |f (x)− L| <ε. The approach is very similar to the one used for sequences. Many of thetechniques used when finding the limit of a sequence will also be used here.

Example 5.1.18 Prove that limx7→∞

1

x= 0.

Let ε > 0 be given. We want to find w > 0 so that x ∈ (w,∞) ∩ D (f) =⇒∣∣∣∣ 1x − 0

∣∣∣∣ < ε. As usual, we begin with the inequality we are trying to prove.∣∣∣∣ 1x − 0

∣∣∣∣ < ε⇐⇒ 1

|x| < ε

Since we are considering the limit as x → ∞, we can restrict ourselves topositive values of x. Thus, the above inequality can be replaced with

1

x< ε

which is equivalent to x >1

ε. Thus we see that given ε > 0, w =

1

εwill work.

Thus, given ε > 0, we have x ∈(

1

ε,∞)∩D (f) =⇒

∣∣∣∣ 1x − 0

∣∣∣∣ < ε.


f (x) = ∞ or that f (x) → ∞ as x → ∞if ∀M > 0, ∃w > 0 : x ∈ (w,∞) ∩D (f) =⇒ f (x) > M .

The above definition says that f (x) can be made arbitrarily large, simplyby taking x large enough.

Example 5.1.20 Prove that limx→∞

x2 =∞.Given M > 0, we must prove that there exists w > 0 such that x ∈ (w,∞) ∩D(x2)

=⇒ x2 > M . Since we are considering the limit as x → ∞, we canrestrict ourselves to x > 0. In this case

x2 > M ⇐⇒ x >√M


Thus, we see that given M > 0, w =√M will work in other words x ∈(√

M,∞)

=⇒ x2 > M .

The remaining definitions are given below.


f (x) = −∞ or that f (x)→ −∞ as x→∞ if ∀M < 0, ∃w > 0 : x ∈ (w,∞) ∩D (f) =⇒ f (x) < M .

Definition 5.1.22 We say that limx→−∞

f (x) = L or that f (x)→ L as x→ −∞if ∀ε > 0, ∃w < 0 : x ∈ (−∞, w) ∩D (f) =⇒ |f (x)− L| < ε.


f (x) = ∞ or that f (x) → ∞ as x →−∞ if ∀M > 0, ∃w < 0 : x ∈ (−∞, w) ∩D (f) =⇒ f (x) > M .


f (x) = −∞ or that f (x) → −∞ as

x→ −∞ if ∀M < 0, ∃w < 0 : x ∈ (−∞, w) ∩D (f) =⇒ f (x) < M .

Remark 5.1.25 In the above definitions, x ∈ (w,∞) can be replaced by x > wand x ∈ (−∞, w) can be replaced by x < w.

One-sided Limits

When we say x → a, we realize that x can approach a from two sides. If xapproaches a from the right, that is if x approaches a and is greater than a, wewrite x→ a+. Similarly, if x approaches a from the left, that is if x approachesa and is less than a, then we write x→ a−.We can rewrite the above definition for one sided limits with little modifica-

tions. We do it for a few of them.

Definition 5.1.26 We say that limx→a+

f (x) = L or that f (x) → L as x → a+

if ∀ε > 0, ∃δ > 0 : 0 < x− a < δ =⇒ |f (x)− L| < ε

Definition 5.1.27 We say that limx→a−

f (x) = L or that f (x) → L as x → a−

if ∀ε > 0, ∃δ > 0 : 0 < a− x < δ =⇒ |f (x)− L| < ε

Example 5.1.28 Prove that limx→0+

1

x=∞.

Given M > 0, we need to prove that there exists δ > 0 such that 0 < x − 0 <

δ =⇒ 1

x> M .

1

x> M ⇐⇒ x <

1

M

Thus, given M > 0, we see that δ =1

Mwill work in other words, we will have

0 < x− 0 <1

M=⇒ 1

x> M .


Theorem 5.1.29 The following two conditions are equivalent

1. limx→a

f (x) = L

2. limx→a+

f (x) = L and limx→a−

f (x) = L

Proof. See problems.

Remark 5.1.30 One way to prove that limx→a

f (x) does not exits is to prove that

the two one-sided limits are not the same or that at least one of them does notexist.

Remark 5.1.31 One sided limits are often used when the definition or behaviorof f changes around the point a at which the limit is being computed. Thiscan happen with piecewise functions when we compute their limit at one of thebreaking points.

We now look at several examples illustrating various techniques used whencomputing limits.

5.1.3 Computing Limits Using the definitions: Examples

Example 5.1.32 Show that limx→3

(4x− 5) = 7

We need to show that given ε > 0, one can find δ such that 0 < |x− 3| < δ =⇒|(4x− 5)− 7| < ε. The strategy is to start with |(4x− 5)− 7| < ε and changethe absolute value to get an inequality with |x− 3|.

|(4x− 5)− 7| < ε⇐⇒ |4x− 12| < ε

⇐⇒ 4 |x− 3| < ε

⇐⇒ |x− 3| < ε

4

So, given ε > 0, δ =ε

4will work.


(x2 + 2

)= 6

We need to show that given ε > 0, one can find δ such that 0 < |x− 2| < δ =⇒∣∣(x2 + 2)− 6∣∣ < ε. ∣∣(x2 + 2

)− 6∣∣ < ε⇐⇒

∣∣x2 − 4∣∣ < ε

⇐⇒ |x− 2| |x+ 2| < ε

Since x2 + 2 is defined for all real numbers, we can assume it is at least definedin (0, 4) (i.e. in (2− h, 2 + h) with h = 2). In this interval, |x+ 2| < 6,therefore,|x− 2| |x+ 2| < 6 |x− 2|. So if we make 6 |x− 2| < ε, the result will

follow. This happens when |x− 2| < ε

6. So, given ε > 0, δ = min

( ε6, 2)will

work.


Remark 5.1.34 In the previous example, the choice of the interval (0, 4) isnot magic. We could have chosen another interval. We want to use an intervalcentered at the point where we are computing the limit. Some readers may thinkwe cheated by only looking at values of x in some intervals. Remember whatwe are trying to achieve. Given an ε > 0, we are finding δ > 0 with certainproperties. As long as we find a δ, we have achieved what we had to achieve.By picking an interval, we simply acknowledge the fact that it is too diffi cult tolook for just any δ, so we restrict our search to a smaller interval.


√x = 2.

We need to show that given ε > 0, one can find δ such that 0 < |x− 4| < δ =⇒|√x− 2| < ε. As before, we start with what we want, and manipulate it until

we get |x− 4| involved.

∣∣√x− 2∣∣ < ε ⇐⇒

∣∣∣∣ (√x− 2) (√x+ 2)√

x+ 2

∣∣∣∣ < ε

⇐⇒∣∣∣∣ x− 4√x+ 2

∣∣∣∣ < ε

⇐⇒ |x− 4|√x+ 2

< ε

Since we are computing the limit as x → 4, we are looking at values of x inan interval of the form (4− h, 4 + h). If we restrict ourselves to h = 2, thenwe are looking at values of x in (2, 6). There,

√x + 2 >

√2 + 2 > 2. Hence,

|x− 4|√x+ 2

<|x− 4|

2. So, if we make

|x− 4|2

< ε that is |x− 4| < 2ε, the result

will follow. So, given ε > 0, δ = min (2ε, 2) will work.

Our next example illustrates how to work with piecewise functions.

Example 5.1.36 Let f (x) =

x if x < 0x2 if 0 < x ≤ 2

8− x if x > 2

1. Prove that limx→0

f (x) = 0.

Since the definition of f changes at 0, we will need to consider one-sidedlimits. We can prove that lim

x→0f (x) = 0 by proving that lim

x→0+f (x) = 0

and limx→0−

f (x) = 0.

• limx→0−

f (x) = 0.

Let ε > 0 be given. We want to prove there exists δ > 0 such that0 < 0 − x < δ ⇐⇒ |f (x)− 0| < ε. If x → 0− then x < 0. In this


case, f (x) = x. Thus, we have:

|f (x)− 0| < ε⇐⇒ |x− 0| < ε

⇐⇒ |x| < ε

⇐⇒ −ε < x < ε

⇐⇒ −ε < x < 0 (since x < 0)

Thus given ε > 0, δ = ε will work.

• limx→0+

f (x) = 0.

Let ε > 0 be given. We want to prove there exists δ > 0 such that0 < x − 0 < δ ⇐⇒ |f (x)− 0| < ε. If x → 0+ then x > 0. In thiscase, f (x) = x2. Thus, we have:

|f (x)− 0| < ε⇐⇒∣∣x2 − 0

∣∣ < ε

⇐⇒ x2 < ε

⇐⇒ |x| <√ε

⇐⇒ 0 < x <√ε (since x > 0)

Thus, given ε > 0, we see that δ =√ε will work.

2. Prove that limx→2

f (x) doe not exist.

As for the previous question, we need to consider one-sided limits. Wewill prove that lim

x→2f (x) doe not exist by proving the one-sided limits are

different. more specifically, we prove that limx→2+

f (x) = 4 and limx→2−

f (x) =

6.

• limx→2−

f (x) = 4.

Let ε > 0 be given. We want to prove there exists δ > 0 such that0 < 2 − x < δ ⇐⇒ |f (x)− 4| < ε. If x → 2− then x < 2. In thiscase, f (x) = x2. Thus, we have:

|f (x)− 4| < ε⇐⇒∣∣x2 − 4

∣∣ < ε

⇐⇒ |x− 2| |x+ 2| < ε

Here, we use a technique similar to one used in an example above.Since we are computing the limit of f (x) as x → 2−, f has to bedefined in an interval of the form (2− h, 2) for some h. We canrestrict ourselves to (0, 2) (this is the case when h = 2). Indeed, fis defined in this interval. Also, in this interval, |x+ 2| < 4 andtherefore

|f (x)− 4| < ε⇐⇒ |x− 2| < ε

4

Thus, we see that given ε > 0, δ = min( ε

4, 2)will work.


• limx→2+

f (x) = 6.

Let ε > 0 be given. We want to prove there exists δ > 0 such that0 < x − 2 < δ ⇐⇒ |f (x)− 6| < ε. If x → 2+ then x > 2. In thiscase, f (x) = 8− x. Thus, we have:

|f (x)− 6| < ε⇐⇒ |8− x− 6| < ε

⇐⇒ |−x+ 2| < ε

⇐⇒ |x− 2| < ε

We see that in this case given ε > 0, δ = ε will work.

Example 5.1.37 Find limx→4

√x− 2

x− 4.

First, we note that if x 6= 4, we have:

√x− 2

x− 4=

√x− 2

x− 4

√x+ 2√x+ 2

=(√x− 2) (

√x+ 2)

(x− 4) (√x+ 2)

=(√x)2 − 4

(x− 4) (√x+ 2)

=x− 4

(x− 4) (√x+ 2)

=1√x+ 2

So that when x→ 4, it seems reasonable to think that√x− 2

x− 4=

1√x+ 2

→ 1

4.

We prove it. Given ε > 0, we need to find δ > 0 such that 0 < |x− 4| < δ =⇒

172 CHAPTER 5. LIMIT OF A FUNCTION∣∣∣∣√x− 2

x− 4− 1

4

∣∣∣∣ < ε.

∣∣∣∣√x− 2

x− 4− 1

4

∣∣∣∣ < ε⇐⇒∣∣∣∣ 1√x+ 2

− 1

4

∣∣∣∣ < ε

⇐⇒∣∣∣∣4−√x− 2

4 (√x+ 2)

∣∣∣∣ < ε

⇐⇒∣∣∣∣ 2−

√x

4 (√x+ 2)

∣∣∣∣ < ε

⇐⇒∣∣∣∣ 2−

√x

4 (√x+ 2)

∣∣∣∣ < ε

⇐⇒∣∣∣∣∣ (2−

√x) (2 +

√x)

4 (√x+ 2)

2

∣∣∣∣∣ < ε

⇐⇒∣∣∣∣∣ (4− x)

4 (√x+ 2)

2

∣∣∣∣∣ < ε

⇐⇒ |4− x|4 (√x+ 2)

2 < ε

Since

|4− x|4 (√x+ 2)

2 =|x− 4|

4 (√x+ 2)

2

<|x− 4|

4x

Since x → 4, we can restrict ourselves to the interval (3, 5) that is an intervalof the form (4− h, 4 + h) with h = 1. In this interval 12 < 4x < 20. Thus, inthis interval

|x− 4|4x

<|x− 4|

12

So, we make|x− 4|

12< ε the result will follow. This will happen if and only if

|x− 4| < 12ε. Thus, given ε > 0, δ = min (1, 12ε) will work. It follows that

limx→4

√x− 2

x− 4=

1

4

The next example illustrates the fact that δ depends not only on ε but alsoon the point at which the limit is being found.

Example 5.1.38 Prove that limx→a

1

x=

1

afor any a ∈ (0,∞).

Let ε > 0 be given. We want to find δ > 0 such that 0 < |x− a| < δ =⇒

5.1. DEFINITIONS AND EXAMPLES 173∣∣∣∣ 1x − 1

a

∣∣∣∣ < ε. We begin with∣∣∣∣ 1x − 1

a

∣∣∣∣ < ε⇐⇒∣∣∣∣a− xax

∣∣∣∣ < ε

⇐⇒ |x− a|ax

< ε

When we are looking at values of x satisfying |x− a| < δ the intent is thatδ be small in other words that be x is close to a. We may then assume that

|x− a| < a

2in which case x >

a

2. Therefore

|x− a|ax

<2 |x− a|a2

. So, if we make

2 |x− a|a2

< ε we will have what we want. This will happen when |x− a| < a2ε

2.

Thus, we see that if δ = min

(a

2,a2ε

2

)then 0 < |x− a| < δ =⇒

∣∣∣∣ 1x − 1

a

∣∣∣∣ < ε.

In particular, we see that δ depends on both ε and a. Of course, one mightargue that we could have done better and found a δ which did not depend ona. We will prove by contradiction that δ must depend on a. Suppose that itdid not, that is for a given ε > 0, the choice of δ does not depend on a. Inparticular, this would work for ε = 1. So, for ε = 1, one can find δ > 0 such

that 0 < |x− a| < δ =⇒∣∣∣∣ 1x − 1

a

∣∣∣∣ < 1. If it works for this δ,it must also work

for any smaller δ. We may then assume that 0 < δ <1

2. Since the choice of δ

is supposed to be independent of a, it should work for a =δ

2. If x = δ, we have

0 < |x− a| =∣∣∣∣δ − δ

2

∣∣∣∣ =

∣∣∣∣δ2∣∣∣∣ < δ

Therefore, we should have

∣∣∣∣ 1x − 1

a

∣∣∣∣ < 1. But∣∣∣∣ 1x − 1

a

∣∣∣∣ =

∣∣∣∣1δ − 2

δ

∣∣∣∣=

1

δ> 1

which is a contradiction.

Example 5.1.39 Let f : R→ R defined by

f (x) =

{1 if x ∈ Q0 if x /∈ Q

Prove that limx→a

f (x) does not exist for any a ∈ R.Fix a ∈ R. We show that if L ∈ R then L cannot be the limit of f (x) as x→ a


by showing that there exists ε > 0 for which no δ will work that is no matterwhich δ we pick, there exists an x satisfying 0 < |x− a| < δ yet |f (x)− L| ≥ ε.Let ε = max {|L− 1| , |L|}. Either ε = |L− 1| or ε = |L|.

case 1: ε = |L− 1|. Since Q is dense in R, one can find x ∈ Q such that forany δ > 0 we have

0 < |x− a| < δ

For such x, |f (x)− L| = |1− L| = ε.

case 2: ε = |L|. Since between any two real numbers there exists an irrationalnumber, for any δ > 0 one can find an irrational number x such that

0 < |x− a| < δ

For such x, |f (x)− L| = |L| = ε.

Conclusion: We have proven that no matter which δ we pick, one can find xsatisfying 0 < |x− a| < δ yet |f (x)− L| ≥ ε. This shows L cannot be alimit of f at a. Since L was arbitrary, lim

x→af (x) does not exist.

5.1.4 Exercises

1. Use the definition of the limit of a function to show that limx→2

√x =√

2.


1

x=

1

2.

3. Discuss the one-sided limits of f (x) = e1x at x = 0.

4. Write a careful proof of the results stated below.

(a) limx→−1

(x3 − 3x

)= 2

(b) limx→3

1

x=

1

3

(c) limx→2

x3 − 8

x2 + x− 6=

12

5

5. Let f (x) =

{x if if x is rationalx2 if if x is irrational

Prove the following:

(a) limx→1

f (x) = 1

(b) limx→2

f (x) does not exist.

6. Prove theorem 5.1.29.


5.1.5 Hints for the Exercises


√x =√

2.

Hint: As the problem says, use the definition. You will also need tomultiply by the conjugate.


1

x=

1

2.

Hint: As in some of the examples, you will need to restrict your search tosome interval around 2, the number x is approaching.

3. Discuss the one-sided limits of f (x) = e1x at x = 0.

Hint: Study carefully the behavior of1

xas x→ 0.

4. Write a careful proof of the results stated below.

(a) limx→−1

(x3 − 3x

)= 2

Hint: As in some of the examples, you will need to restrict yoursearch to some interval around 1, the number x is approaching.

(b) limx→3

1

x=

1

3Hint: Similar hint as above.

(c) limx→2

x3 − 8

x2 + x− 6=

12

5Hint: Similar hint as above.

5. Let f (x) =

{x if if x is rationalx2 if if x is irrational

Prove the following:

(a) limx→1

f (x) = 1

Hint: Using the definition of limits, write what you have to prove,and use techniques similar to the previous problems and the fact thatboth rationals and irrationals are dense in R.

(b) limx→2


Hint: Do a proof by contradiction.

6. Prove theorem 5.1.29.Hint: Just use the definitions of the various concepts involved.


5.2 Limit Theorems

5.2.1 Operations with Limits

The same theorem we proved for sequences also hold for functions. We list thetheorem, and leave its proof as an exercise.

Theorem 5.2.1 Assuming that limx→a

f (x) and limx→a

g (x) exist, the following re-

sults are true:

1. limx→a

(f (x)± g (x)) = limx→a

f (x)± limx→a

g (x)

2. limx→a

(f (x) g (x)) =(

limx→a

f (x))(

limx→a

g (x))

3. limx→a

f (x)

g (x)=

limx→a

f (x)

limx→a

g (x)as long as lim

x→ag (x) 6= 0

4. limx→a|f (x)| =

∣∣∣ limx→a

f (x)∣∣∣

5. If f (x) ≥ 0, then limx→a

f (x) ≥ 0

6. If f (x) ≥ g (x) then limx→a

f (x) ≥ limx→a

g (x)

7. If f (x) ≥ 0, then limx→a

√f (x) =

√limx→a

f (x)

Remark 5.2.2 The above results also hold when the limits are taken as x →±∞.

Remark 5.2.3 All the techniques learned in Calculus can be used here. Thesetechniques include factoring, multiplying by the conjugate.

We look at a few examples to refresh the reader’s memory of some standardtechniques.


x2 − 6x+ 8

x− 4Here, we note that both the numerator and denominator are approaching 0 asx → 4. Since both the numerator and denominator are polynomials, we knowwe can factor x− 4.

limx→4

x2 − 6x+ 8

x− 4= lim

x→4

(x− 4) (x− 2)

x− 4= lim

x→4(x− 2)

= 2

5.2. LIMIT THEOREMS 177


√x− 1

x− 1Here, we note that both the numerator and denominator are approaching 0 asx → 1. The fraction also contains a radical. A standard technique is to try tomultiply by the conjugate.

limx→1

√x− 1

x− 1= lim

x→1

(√x− 1) (

√x+ 1)

(x− 1) (√x+ 1)

= limx→1

x− 1

(x− 1) (√x+ 1)

= limx→1

1√x+ 1

1

2

Example 5.2.6 Find limx→∞

x2 + 5x+ 1

2x2 − 10When taking the limit as x→ ±∞ of a rational function, a standard techniqueis to factor the term of highest degree from both the numerator and denominator.

limx→∞

x2 + 5x+ 1

2x2 − 10= limx→∞

x2(

1 +5

x+

1

x2

)x2(

2− 10

x2

)

As x→∞, 1 +5

x+

1

x2→ 1 and 2− 10

x2→ 2, so

limx→∞

x2(

1 +5

x+

1

x2

)x2(

2− 10

x2

) = limx→∞

x2

2x2

=1

2

Remark 5.2.7 The above example is a special case of a more general result wegive below.

Theorem 5.2.8 The limit as x→ ±∞ of a rational function is the limit of thequotient of the terms of highest degree.Proof. See problems.

Example 5.2.9 Find limx→∞

3x2 + 1

4x3 + 2x+ 1

From the above theorem, we see that limx→∞

3x2 + 1

4x3 + 2x+ 1= limx→∞

3x2

4x3= limx→∞

3

4x=

0.


5.2.2 Elementary Theorems

Theorems similar to those studied for sequences hold. We will leave the proofof most of these as an exercise.

Theorem 5.2.10 If the limit of a function exists, then it is unique.Proof. See exercises at the end of this section.

The next theorem relates the notion of limit of a function with the notionof limit of a sequence.

Theorem 5.2.11 Suppose that limx→a

f (x) = L and that {xn} is a sequence ofpoints such that lim

n→∞xn = a, xn 6= a∀n. Put yn = f (xn). Then, lim

n→∞yn = L

Proof. Let ε > 0 be given. Since limx→a

f (x) = L, we can find δ such that

0 < |x− a| < δ =⇒ |f (x)− L| < ε. Since limn→∞

xn = a, we can find N such

that n ≥ N =⇒ |xn − a| < δ. But in this case, it follows that |f (xn)− L| < ε.

The converse of this theorem is also true.

Theorem 5.2.12 If f is defined in a deleted neighborhood of a such that f (xn)→L for every sequence {xn} such that xn → a, then lim

x→af (x) = L.

Proof. See problems at the end of this section.

Like for sequences, if a function has a limit at a point, then it is bounded.However, we need to be a little bit more careful here. If lim

x→af (x) = L, then

we are only given information about the behavior of f close to a. Therefore, wecan only draw conclusions about what happens to f as long as x is close to a.

Theorem 5.2.13 If limx→a

f (x) = L, then there exists a deleted neighborhood of

a in which f is bounded.Proof. We can find δ such that 0 < |x− a| < δ =⇒ |f (x)− L| < 1. By thetriangle inequality, we have for such x′s

|f (x)| − |L| ≤ ||f (x)| − |L||≤ |f (x)− L|< 1

It follows that|f (x)| < 1 + |L|

Therefore, f is bounded by 1 + |L| in (a− δ, a+ δ).

Theorem 5.2.14 Suppose that f (x) ≤ g (x) ≤ h (x) in a deleted neighborhoodof a and lim

x→af (x) = lim

x→ah (x) = L then lim

x→ag (x) = L.

Proof. We do a direct proof here. In section 5.2.3 , we’ll see another proof.Suppose the above inequality holds in (a− δ1, a+ δ1) for some δ1 > 0. Let ε > 0


be given. Choose δ2 such that 0 < |x− a| < δ2 =⇒ ε−L < f (x) < ε+L. Chooseδ3 such that 0 < |x− a| < δ3 =⇒ ε−L < h (x) < ε+L. Let δ = min (δ1, δ2, δ3).Then, if 0 < |x− a| < δ, we have

ε− L < f (x) ≤ g (x) ≤ h (x) < ε+ L

⇐⇒ ε− L < g (x) < ε+ L

⇐⇒ |g (x)− L| < ε

Therefore, limx→a

g (x) = L.


x2 sin

(1

x

)= 0.

5.2.3 Relationship Between the Limit of a Function andthe Limit of a Sequence

Theorems 5.2.11 and 5.2.12 can be summarized in the theorem below.

Theorem 5.2.16 Let f be a function of one real variable defined in a deletedneighborhood of a real number a. The following conditions are equivalent.

1. limx→a

f (x) = L

2. For every sequence {xn} such that xn 6= a and xn → a we have limn→∞

f (xn) =

L.

Proof. We prove both directions.

• (1 =⇒ 2). We assume that limx→a

f (x) = L. Let {xn} be a sequence suchthat xn 6= a and xn → a. Then, the conclusion follows from theorem5.2.11.

• (2 =⇒ 1). This is theorem 5.2.12.

This theorem provides the link between the limit of a function and the limitof a sequence. Using this theorem, we can prove the theorems about the limitof a function by using their counterpart for sequences. We illustrate this withanother version of the proof of the squeeze theorem.

Theorem 5.2.17 Suppose that f (x) ≤ g (x) ≤ h (x) in a deleted neighborhoodof a and lim

x→af (x) = lim

x→ah (x) = L then lim

x→ag (x) = L.

Proof. To show that limx→a

g (x) = L, we need to show that if xn is any sequence

which converges to a, then g (xn) is a sequence which converges to L. Let xnbe a sequence such that xn → a, xn 6= a. Then, f (xn) → L and h (xn) → L.Since f (xn) ≤ g (xn) ≤ h (xn) we can apply the squeeze theorem for sequencesto conclude that g (xn)→ L.


Theorem 5.2.16 also gives us a convenient way to show that a limit of afunction does not exist. We summarize how in the following corollary.

Corollary 5.2.18 Let f be a function of one real variable defined in a deletedneighborhood of a real number a. Then, lim

x→af (x) does not exist if either of the

following conditions holds:

1. There exist sequences (xn) and (yn) with xn 6= a and yn 6= a such thatlimxn = lim yn = a but lim

n→∞f (xn) 6= lim

n→∞f (yn).

2. There exists a sequence (xn) with xn 6= a such that limxn = a but thesequence f (xn) diverges.

We illustrate this corollary with an example.

Example 5.2.19 Consider f : R� {0} → R given by f (x) = sin

(1

x

). Show

limx→0


Consider the sequences xn =1

2πnand yn =

1

2πn+π

2

. Then, clearly, limxn =

lim yn = 0. Yet, f (xn) = sin (2πn) = 0 thus limn→∞

f (xn) = 0 but f (yn) =

sin(

2πn+π

2

)= 1 thus lim

n→∞f (yn) = 1.

5.2.4 Exercises

1. Prove theorem 5.2.8.

2. Prove theorem 5.2.10. Do both a direct proof and a proof using sequences.

3. Prove theorem 5.2.1. Do both a direct proof and a proof using sequences.

4. Evaluate the following limits:

(a) limx→−2

x2 − 4

x+ 2

(b) limx→3

x3 − 27

x− 3

(c) limx→1

xn − 1

x− 1where n is a positive integer.

(d) limx→1

xn − 1

xm − 1

(e) limx→6

√x− 2− 2

x− 6

5. Assuming you know that limx→0

sinx

x= 1, compute the limits below:


(a) limx→0

sin 2x

x

(b) limx→0

sin 3x

sin 5x

(c) limx→0

x

tanx

(d) limx→0

sinx√x

6. Evaluate limx→0

√(a+ bx) (c+ dx)−

√ac

x

7. Prove Theorem 5.2.12.

8. Prove that if limx→a

f (x) > 0 then there exists δ > 0 such that f (x) > 0 for

every x ∈ (a− δ, a+ δ).

9. We say that a function f : I → R where I ⊆ R is Lipschitz providing thereexists a constant K > 0 such that

|f (x)− f (y)| ≤ K |x− y|

for x, y ∈ I.

(a) Give an example of a Lipschitz function. You must show that thefunction in your example satisfies the required condition.

(b) Prove rigorously that if f is Lipschitz, then limx→a

f (x) = f (a).


5.2.5 Hints for the Exercises

1. Prove theorem 5.2.8.Hint: Write the expression of a typical rational function, then factor theterm of highest degree from both the numerator and denominator.

2. Prove theorem 5.2.10.Hint: The proof is similar to the same result for sequences.

3. Prove theorem 5.2.1.Hint: Rewrite the result to prove using theorems 5.2.11 and 5.2.12, thenuse similar results for sequences.

4. Evaluate the following limits:

(a) limx→−2

x2 − 4

x+ 2Hint: Use simple algebra to factor some terms.

(b) limx→3

x3 − 27

x− 3Hint: Use simple algebra to factor some terms.

(c) limx→1

xn − 1

x− 1where n is a positive integer.

Hint: Use simple algebra to factor some terms, remembering thatxn − 1 = (x− 1)

(1 + x+ x2 + ...+ xn−1

)(d) lim

x→1

xn − 1

xm − 1Hint: Use simple algebra to factor some terms, remembering thatxn − 1 = (x− 1)

(1 + x+ x2 + ...+ xn−1

)(e) lim

x→6

√x− 2− 2

x− 6Hint: Use the conjugate.

5. Assuming you know that limx→0

sinx

x= 1, compute the limits below:

(a) limx→0

sin 2x

xHint: Use the fact that x→ 0⇐⇒ 2x→ 0.

(b) limx→0

sin 3x

sin 5xHint: Remembering the previous problem, rewrite this problem so it

looks likesinx

xso you can use the given result.

(c) limx→0

x

tanxHint: Use the definition of tanx.


(d) limx→0

sinx√x

Hint: In order to use the given result, you need to rewrite this sothat there is an x in the denominator.

6. Evaluate limx→0

√(a+ bx) (c+ dx)−

√ac

xHint: Use the conjugate.

7. Prove Theorem 5.2.12.Hint: Using the definition of limits, try a proof by contradiction.

8. Prove that if limx→a

f (x) > 0 then there exists δ > 0 such that f (x) > 0 for

every x ∈ (a− δ, a+ δ).Hint: Using the definition of limits, pick an appropriate ε to have thedesired result.

9. We say that a function f : I → R where I ⊆ R is Lipschitz providing thereexists a constant K > 0 such that

|f (x)− f (y)| ≤ K |x− y|

for x, y ∈ I.

(a) Give an example of a Lipschitz function. You must show that thefunction in your example satisfies the required condition.Hint: Try to understand the meaning of this condition. To help you

do that, think of the geometric meaning of the quantityf (x)− f (y)

x− y .

(b) Prove rigorously that if f is Lipschitz, then limx→a

f (x) = f (a).

Hint: Simply use the definition of limit.

Bibliography

[B] Bartle, G. Robert, The elements of Real Analysis, Second Edition, JohnWiley & Sons, 1976.

[C1] Courant, R., Differential and Integral Calculus, Second Edition, Volume 1,Wiley, 1937.

[C2] Courant, R., Differential and Integral Calculus, Second Edition, Volume 2,Wiley, 1937.

[DS] Dangello, Frank & Seyfried Michael, Introductory Real Analysis, HoughtonMiffl in, 2000.

[F] Fulks, Watson, Advanced Calculus, an Introduction to Analysis, Third Edi-tion, John Wiley & Sons, 1978.

[GN] Gaskill, F. Herbert & Narayanaswami P. P., Elements of Real Analysis,Prentice Hall, 1998.

[LL] Lewin, J. & Lewin, M., An Introduction to Mathematical Analysis, SecondEdition, McGraw-Hill, 1993.

[MS] Stoll, Manfred, Introduction to Real Analysis, Second Edition, Addison-Wesley Higher Mathematics, 2001.

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