Chapter 5

Limits of a Function
5.1
5.1.1

Denitions and Examples

Overview

Unlike for sequences, there are many possibilities for the limit of a function. In
this section, we will investigate the following limits: lim f (x) and lim f (x).
x! 1

x!a

Given a function y = f (x), we are studying how f behaves near a point a or

near innity. The questions we are trying to answer are:
1. As x is getting closer and closer to a, how is y = f (x) behaving? Is it
getting closer to a number as well? Is it getting arbitrary large (in absolute
value)? Is it not following any pattern?
2. Same question if x is approaching 1 or

1.

In more general terms, we are asking the question: if x is following a certain

pattern, is f (x) also following a pattern and if yes, which pattern?
In order to be able to evaluate lim f (x), f must be dened in a deleted neighx!a

borhood of a that is f must be dened in an interval of the form (a h; a + h)

for some positive number h;except maybe at x = a.
When we say lim f (x) = L, we mean that f (x) can be made as close
x!a
as we want from L simply by taking x close enough to a. Or, in terms of
neighborhoods, we have the following general denition for a limit.
Denition 362 We say that lim f (x) = L or that f (x) ! L as x ! a if for
x!a
every neighborhood V of L, one can nd a deleted neighborhood U of a such that
x 2 U =) f (x) =) L.
This denition can be adapted to limits at a nite point or at innity as
well as when the limit is nite or innite. There are three possibilities for x, we
can have x ! a, x ! 1 and x ! 1. For each case, we can have f (x) ! L,
139

140

CHAPTER 5. LIMITS OF A FUNCTION

f (x) ! 1 and f (x) ! 1. Hence, we have a total of nine denitions. They

can all be derived from the above denition simply by remembering that a
neighborhood of a nite point a is an interval of the form (a
; a + ) and a
neighborhood of innity is an interval of the form (w; 1) for some w 2 R.

5.1.2

Limit at a nite point

Denition 363 We say that lim f (x) = L or that f (x) ! L as x ! a if

x!a

8 > 0, 9 > 0 : 0 < jx

aj <

=) jf (x)

Lj < .

jx aj 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 364 Prove that lim (x + 5) = 7.
x!2

Given > 0, we must prove that there exists

jx + 5 7j < . Let > 0 be given.
jx + 5
Thus we see that
jx + 5 7j < .

> 0 such that 0 < jx

7j < () jx

2j <

=)

2j <

=)

2j <

will work. Indeed, given

> 0, 0 < jx

Remark 365 Of course, this was a very easy example to illustrate how this
kind of problem is addressed. In general, it will take more work to nd given
> 0. Many of the techniques used for sequences will also be used here. Some
will be illustrated below when we look at more challenging examples.
Remark 366 Saying that 0 < jx aj <
is the same as saying that x 2
(a
; a + ) and x 6= a. Similarly, saying that jf (x) Lj < is the same as
saying that f (x) 2 (L
; L + ).
Remark 367 In order to understand how to adapt this denition to cases involving 1 such as in the case when L = 1 or also when x ! 1, it is important
to understand the notion of neighborhood of 1. A neighborhood of 1 is an
interval of the form (w; 1). Similarly, an neighborhood of 1 is an interval of
the form ( 1; w).
Denition 368 We say that lim f (x) = 1 or that f (x) ! 1 as x ! a if
x!a

8M > 0, 9 > 0 : 0 < jx

aj <

=) f (x) > M .
1

2 = 1.
(x 1)
Given M > 0, we must prove that there exists

Example 369 Prove that lim

x!1

> 0 such that 0 < jx

1j <

5.1. DEFINITIONS AND EXAMPLES

=)

1
(x

1)

141

> M.
1
(x

Thus, given M > 0,

1)

>

M () (x

()

()

jx

1) <

1
M

1
<x
M

1
1< p
M
1
1j < p
M

1
=p
will work.
M

Denition 370 We say that lim f (x) =

x!a

8M < 0, 9 > 0 : 0 < jx

aj <

1 or that f (x) !

=) f (x) < M .

1 as x ! a if

Remark 371 If we nd a which works, then every 0 < will also work.
Therefore, it is always possible to impose certain conditions on such as saying
that we are looking for less than a certain number h, thus restricting our search
to an interval of the form (a h; a + h). In this interval, if we call 0 the value
we found, then = min h; 0 .
Remark 372 In the last two denition, the vertical line x = a is a vertical
asymptote for the graph of y = f (x).
Remark 373 In the denition of a limit, it is implied that a is a limit point of
D (f ) that is for every > 0 the interval (a
; a + ) contains points of D (f )
other than a. If this is not the case, then for small enough, there may not
exist any x satisfying 0 < jx aj < . In this case, the concept of a limit has
no meaning.
Remark 374 In the denition of a limit, a does not have to be in the domain
of f . It only needs to be a limit point of D (f ).
Remark 375 In the denition of a limit, depends obviously on . It may also
depend on the point a as illustrated by example 398.
Remark 376 When we say that the limit of a function exists, we mean that
it exists and is nite. When the limit is innite, it does not exist in the sense
that it is not a number. However, we know what the function is doing, it is
approaching 1.
Remark 377 There are several situations under which a limit will fail to exist.
1
as x ! 0.
x
2. The function may oscillate unboundedly like x sin x as x ! 1.
1. The function may oscillate boundedly like in f (x) = sin

3. The function may grow without bounds like

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

1
as x ! 0.
x2

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CHAPTER 5. LIMITS OF A FUNCTION

Limit at innity
In order to be able to evaluate lim f (x), f must be dened for large x. In
x!1

other words, we must have D (f ) \ (w; 1) 6= ? for every w 2 R. In the case we

want to evaluate lim f (x), then we must have D (f ) \ ( 1; w) 6= ? for every
x! 1

w 2 R. We then have the following denitions (some of the denitions will be

accompanied with easy examples to illustrate the concept being dened):

Denition 378 We say that lim f (x) = L or that f (x) ! L as x ! 1 if

x!1

8 > 0, 9w > 0 : x 2 (w; 1) \ D (f ) =) jf (x)

Lj <

jf (x) Lj represents the distance between f (x) and L. The above statement simply says that f (x) can be made as close as one wants to L, simply by
taking x large enough. Graphically, this simply says that the line y = L is a
horizontal asymptote for the graph of y = f (x).
To prove that a number f (x) approaches L as x ! 1, given > 0, one has
to prove that w > 0 can be found so that x 2 (w; 1) \ D (f ) =) jf (x) Lj <
. The approach is very similar to the one used for sequences. Many of the
techniques used when nding the limit of a sequence will also be used here.
1
= 0.
x7!1 x
> 0 be given. We want to nd w > 0 so that x 2 (w; 1) \ D (f ) =)

Example 379 Prove that lim

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

0 < ()

1
<
jxj

Since we are considering the limit as x ! 1, we can restrict ourselves to

1
positive values of x. Thus, the above inequality can be replaced with
<
x
1
1
which is equivalent to x > . Thus we see that given > 0, w = will work.
Thus, given

> 0, we have x 2

; 1 \ D (f ) =)

1
x

0 < .

Denition 380 We say that lim f (x) = 1 or that f (x) ! 1 as x ! 1 if

x!1

8M > 0, 9w > 0 : x 2 (w; 1) \ D (f ) =) f (x) > M .

The above denition says that f (x) can be made arbitrarily large, simply
by taking x large enough.
Example 381 Prove that lim x2 = 1.
x!1

Given M > 0, we must prove that there exists w > 0 such that x 2 (w; 1) \
D x2 =) x2 > M . Since we are considering the limit as x ! 1, we can
restrict ourselves to x > 0. In this case
p
x2 > M () x > M

5.1. DEFINITIONS AND EXAMPLES

Thus, we see that given M > 0, w =
p
M ; 1 =) x2 > M .

143
p

M will work in other words x 2

Denition 382 We say that lim f (x) =

1 or that f (x) !

x!1

if 8M < 0, 9w > 0 : x 2 (w; 1) \ D (f ) =) f (x) < M .

1 as x ! 1

Denition 383 We say that lim f (x) = L or that f (x) ! L as x !

x! 1

8 > 0, 9w < 0 : x 2 ( 1; w) \ D (f ) =) jf (x)

Denition 384 We say that

Lj < .

1 if

lim f (x) = 1 or that f (x) ! 1 as x !

x! 1

if 8M > 0, 9w < 0 : x 2 ( 1; w) \ D (f ) =) f (x) > M .

Denition 385 We say that

lim f (x) =

x! 1

1 or that f (x) !

1 if 8M < 0, 9w < 0 : x 2 ( 1; w) \ D (f ) =) f (x) < M .

1 as x !

One-sided Limits
When we say x ! a, we realize that x can approach a from two sides. If x
approaches a from the right, that if x approaches a and is greater than a, we
write x ! a+ . Similarly, if x approaches a from the left, that is if x approaches
a and is less than a, then we write x ! a .
We can rewrite the above denition for one sided limits with little modications. We do it for a few of them.
Denition 386 We say that lim+ f (x) = L or that f (x) ! L as x ! a+ if
x!a

8 > 0, 9 > 0 : 0 < x

a<

=) jf (x)

Lj <

Denition 387 We say that lim f (x) = L or that f (x) ! L as x ! a

x!a

8 > 0, 9 > 0 : 0 < a

x<

=) jf (x)

if

Lj <

1
= 1.
x
Given M > 0, we need to prove that there exists > 0 such that 0 < x 0 <
1
=) > M .
x
1
1
> M () x <
x
M
1
Thus, given M > 0, we see that =
will work in other words, we will have
M
1
1
0<x 0<
=) > M .
M
x
Example 388 Prove that lim

x!0+

Theorem 389 The following two conditions are equivalent

1. lim f (x) = L
x!a

2. lim f (x) = L and lim f (x) = L

x!a+

x!a

144

CHAPTER 5. LIMITS OF A FUNCTION

Proof. See problems.

Remark 390 One way to prove that lim f (x) does not exits is to prove that
x!a
the two one-sided limits are not the same or that at least one of them does not
exist.
Remark 391 One sided limits are often used when the denition or behavior
of f changes around the point a at which the limit is being computed. This
can happen with piecewise functions when we compute their limit at one of the
breaking points.
We now look at several examples illustrating various techniques used when
computing limits.

5.1.3

Computing Limits Using the denitions: Examples

Example 392 Show that lim (4x

5) = 7

x!3

We need to show that given > 0, one can nd such that 0 < jx 3j < =)
j(4x 5) 7j < . The strategy is to start with j(4x 5) 7j < and change
the absolute value to get an inequality with jx 3j.
j(4x

5)

7j < () j4x
() 4 jx
() jx

So, given

> 0,

12j <

3j <

3j <

will work.

Example 393 Show that lim x2 + 2 = 6

x!2

We need to show that given

x2 + 2
6 < .

> 0, one can nd

x2 + 2

6 < () x2
() jx

such that 0 < jx

2j <

=)

4 <

2j jx + 2j <

Since x2 +2 is dened for all real numbers, we can assume it is at least dened in
(0; 4) (i.e. in (2 h; 2 + h) with h = 2). In this interval, jx + 2j < 6, therefore,
jx
So, given

> 0,

= min

2j <

; 2 will work.

Remark 394 In the previous example, the choice of the interval (0; 4) is not
magic. We could have chosen another interval. We want to use an interval
centered at the point where we are computing the limit. Some readers may think

5.1. DEFINITIONS AND EXAMPLES

145

we cheated by only looking at values of x in some intervals. Remember what

we are trying to achieve. Given an > 0, we are nding > 0 with certain
properties. As long as we nd a , we have achieved what we had to achieve.
By picking an interval, we simply acknowledge the fact that it is too di cult to
look for just any , so we restrict our search to a smaller interval.
p
Example 395 Show that lim x = 2.
x!4

We
p need to show that given > 0, one can nd such that 0 < jx 4j < =)
j x 2j < . As before, we start with what we want, and manipulate it until
we get jx 4j involved.
p
p
p
( x 2) ( x + 2)
p
<
x 2
<
()
x+2
x 4
p
()
<
x+2
jx 4j
() p
<
x+2
Since we are computing the limit as x ! 4, we are looking at values of x in
an interval of the form (4 h; 4 + h). If we restrict
ourselves
to h = 2, then
p
p
we are looking at values of x in (2; 6). There, x + 2 > 2 + 2 > 2. Hence,
jx 4j
jx 4j
jx 4j
p
<
. So, if we make
< that is jx 4j < 2 , the result
2
2
x+2
will follow. So, given > 0, = min (2 ; 2) will work.
Our next example illustrates how
8
< x
x2
Example 396 Let f (x) =
:
8 x

to work with piecewise functions.

if
if
if

x<0
0<x 2
x>2

1. Prove that lim f (x) = 0.

x!0

Since the denition of f changes at 0, we will need to consider one-sided

limits. We can prove that lim f (x) = 0 by proving that lim+ f (x) = 0
x!0

x!0

and lim f (x) = 0.

x!0

lim f (x) = 0.

x!0

Let > 0 be given. We want to prove there exists > 0 such that
0 < 0 x < () jf (x) 0j < . If x ! 0 then x < 0. In this
case, f (x) = x. Thus, we have:
jf (x)

0j

<
()

()

() jx

()
Thus given

> 0,

0j <

jxj <

will work.

<x<
< x < 0 (since x < 0)

146

CHAPTER 5. LIMITS OF A FUNCTION

lim f (x) = 0.

x!0+

Let > 0 be given. We want to prove there exists > 0 such that
0 < x 0 < () jf (x) 0j < . If x ! 0+ then x > 0. In this
case, f (x) = x2 . Thus, we have:
jf (x)

0j

() x2

<
()

x <

()

jxj <

p
0<x<
(since x > 0)
p
> 0, we see that =
will work.
()

Thus, given

0 <

2. Prove that lim f (x) doe not exist.

x!2

As for the previous question, we need to consider one-sided limits. We

will prove that lim f (x) doe not exist by proving the one-sided limits are
x!2

di erent. more specically, we prove that lim+ f (x) = 4 and lim f (x) =
x!2

6.

x!2

lim f (x) = 4.

x!2

Let > 0 be given. We want to prove there exists > 0 such that
0 < 2 x < () jf (x) 4j < . If x ! 2 then x < 2. In this
case, f (x) = x2 . Thus, we have:
jf (x)

4j

() x2

<
()

jx

4 <

2j jx + 2j <

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 be
dened in an interval of the form (2 h; 2) for some h. We can
restrict ourselves to (0; 2) (this is the case when h = 2). Indeed, f
is dened in this interval. Also, in this interval, jx + 2j < 4 and
therefore
jf (x) 4j < () jx 2j <
4
Thus, we see that given

> 0,

= min

; 2 will work.

lim f (x) = 6.

x!2+

Let > 0 be given. We want to prove there exists > 0 such that
0 < x 2 < () jf (x) 6j < . If x ! 2+ then x > 2. In this
case, f (x) = 8 x. Thus, we have:
jf (x)

6j

<
()
()

We see that in this case given

() j8 x
j x + 2j <
jx

2j <

> 0,

6j <

will work.

5.1. DEFINITIONS AND EXAMPLES

147

x 2
.
x!4 x
4
First, we note that if x 6= 4, we have:
Example 397 Find lim

p
p
x 2
x 2 x+2
p
=
x 4
x 4 x+2
p
p
( x 2) ( x + 2)
p
=
(x 4) ( x + 2)
p 2
( x)
4
p
=
(x 4) ( x + 2)
x 4
p
=
(x 4) ( x + 2)
1
=p
x+2

We
p prove it. Given
x 2 1
< .
x 4
4
p

> 0, we need to nd

x 2
x 4

1
4

x 2
1
1
! .
= p
x 4
4
x+2
> 0 such that 0 < jx 4j < =)

So that when x ! 4, it seems reasonable to think that

()

1
1
() p
<
x+2 4
p
4
x 2
p
<
4 ( x + 2)
p
2
x
p
<
4 ( x + 2)
p
2
x
p
<
4 ( x + 2)
p
p
(2
x) (2 + x)
<
p
2
4 ( x + 2)

()

(4 x)
p
2 <
4 ( x + 2)

<
()
()
()

()

j4 xj
p
2 <
4 ( x + 2)

Since
j4 xj
p
2
4 ( x + 2)

=
<

jx 4j
p
2
4 ( x + 2)
jx 4j
4x

148

CHAPTER 5. LIMITS OF A FUNCTION

Since x ! 4, we can restrict ourselves to the interval (3; 5) that is an interval

of the form (4 h; 4 + h) with h = 1. In this interval 12 < 4x < 20. Thus, in
this interval
jx 4j
jx 4j
<
4x
12
4j
< the result will follow. This will happen if and only if
12
jx p
4j < 12 . Thus, given > 0, = min (1; 12 ) will work. It follows that
x 2
1
lim
=
x!4 x
4
4

So, we make

jx

The next example illustrates the fact that depends not only on
on the point at which the limit is being found.
1
1
= for any a 2 (0; 1).
x
a
> 0 be given. We want to nd > 0 such that 0 < jx
1
< . We begin with
a

but also

Example 398 Prove that lim

x!a

Let
1
x

1
x

1
a

<
()

()
jx

ax

aj
ax

aj <

=)

<

<

When we are looking at values of x satisfying jx aj < the intent is that

be small in other words that be x is close to a. We may then assume that
a
a
jx aj
2 jx aj
jx aj < in which case x > . Therefore
<
. So, if we make
2
2
ax
a2
2 jx aj
a2
< we will have what we want. This will happen when jx aj <
.
2
a
2
2
a a
1
1
Thus, we see that if = min
;
< .
then 0 < jx aj < =)
2 2
x a
In particular, we see that depends on both and a. Of course, one might
argue that we could have done better and found a which did not depend on
a. We will prove by contradiction that must depend on a. Suppose that it
did not, that is for a given > 0, the choice of does not depend on a. In
particular, this would work for = 1. So, for = 1, one can nd > 0 such
1
1
< 1. If it works for this ;it must also work
that 0 < jx aj < =)
x a
1
for any smaller . We may then assume that 0 < < . Since the choice of
2
is supposed to be independent of a, it should work for a = . If x = , we have
2
0 < jx

aj =

<

5.1. DEFINITIONS AND EXAMPLES

Therefore, we should have

1
x

149

1
< 1. But
a
1
x

1
a

=
=

> 1
which is a contradiction.
Example 399 Let f : R ! R dened by
f (x) =

1
0

if x 2 Q
if x 2
=Q

Prove that lim f (x) does not exist for any a 2 R.

x!a

Fix a 2 R. We show that if L 2 R then L cannot be the limit of f (x) as x ! a

by showing that there exists > 0 for which no will work that is no matter
which we pick, there exists an x satisfying 0 < jx aj < yet jf (x) Lj
.
Let = max fjL 1j ; jLjg. Either = jL 1j or = jLj.
case 1: = jL 1j. Since Q is dense in R, one can nd x 2 Q such that for
any > 0 we have
0 < jx aj <
For such x, jf (x)

Lj = j1

Lj = .

case 2: = jLj. Since between any two real numbers there exists an irrational
number, for any > 0 one can nd an irrational number x such that
0 < jx
For such x, jf (x)

aj <

Lj = jLj = .

Conclusion: We have proven that no matter which we pick, one can nd x

satisfying 0 < jx aj < yet jf (x) Lj
. This shows L cannot be a
limit of f at a. Since L was arbitrary, lim f (x) does not exist.
x!a

5.1.4

Exercises

1. Use the denition of the limit of a function to show that lim

x!2

x=

1
1
= .
x!2 x
2

2. Use the denition of the limit of a function to show that lim

1

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

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

2.

150

CHAPTER 5. LIMITS OF A FUNCTION

(a) lim

x! 1

x3

3x = 2

1
1
=
x
3
3
x
8
12
(c) lim 2
=
x!2 x + x
6
5

(b) lim

x!3

5. Let f (x) =

x
x2

if
if

if x is rational
Prove the following:
if x is irrational

(a) lim f (x) = 1

x!1

(b) lim f (x) does not exist.

x!2

6. Prove theorem 389.