LIMIT OF A FUNCTION

Lecture Notes

John Mark B. Gildo

jmbgildo@mcm.edu.ph

December 17, 2019

“Mathematics is a place where you can do things

which you can’t do in the real world.”
Marcus du Sautoy

Contents
1 Learning Competencies 1

2 Introduction 1

3 Limit of a function 1
3.1 Table of Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
3.2 Looking at the graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3.2.1 Test your knowledge! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 The Limit of a Function at c versus the Value of the Function at c 10

4.0.1 Test your knowledge! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Illustration of Limit Theorems 12

5.1 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
5.1.1 Test your knowledge! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6 Limits of Polynomial, Rational, and Radical Functions 17

6.1 Limits of polynomial functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
6.2 Limits of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
6.3 Limit of Radical Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
6.4 Infinite limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
6.4.1 Test your knowledge! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

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1 Learning Competencies
1. Illustrate the limit of a function using a table of values and the graph of the function;

2. Distinguish between lim f (x) and f (c);

x→c

3. Illustrate the limit theorems; and

4. Apply the limit theorems in evaluating the limit of algebraic functions (polynomial, rational, and radical).

2 Introduction
Limits are the backbone of calculus, and calculus is called the Mathematics of Change. The study of limits is
necessary in studying change in great detail. The evaluation of a particular limit is what underlies the formulation
of the derivative and the integral of a function.

For starters, imagine that you are going to watch a basketball game. When you choose seats, you would
want to be as close to the action as possible. You would want to be as close to the players as possible and have
the best view of the game, as if you were in the basketball court yourself. Take note that you cannot actually
be in the court and join the players, but you will be close enough to describe clearly what is happening in the game.

This is how it is with limits of functions. We will consider functions of a single variable and study the behavior
of the function as its variable approaches a particular value (a constant). The variable can only take values very,
very close to the constant, but it cannot equal the constant itself. However, the limit will be able to describe
clearly what is happening to the function near that constant.

3 Limit of a function
Limit
Definition 3.1. Consider a function f of a single variable x. Consider a constant c which the variable x
will approach (c may or may not be in the domain of f ). The limit, to be denoted by L, is the unique real
value that f (x) will approach as x approaches c. In symbols, we write this process as

lim f (x) = L.
x→c

This is read as, “The limit of f (x) as x approaches c is L.”

3.1 Table of Values

To illustrate, let us consider
lim (1 + 3x).
x→2

Here, f (x) = 1 + 3x, and the constant c, which x will approach, is 2. To evaluate the given limit, we will make
use of a table to help us keep track of the effect that the approach of x toward 2 will have on f (x). Of course, on
the number line, x may approach 2 in two ways: through values on its left and through values on its right. First
consider approaching 2 from its left or through values less than 2. Remember that the values to be chosen should

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3.1 Table of Values Page 2

be close to 2.

x f (x)
1 4
1.4 5.2
1.7 6.1
1.9 6.7
1.95 6.85
1.997 6.991
1.9999 6.9997
1.9999999 6.9999997

Now we consider approaching 2 from its right or through values greater than but close to 2.

x f (x)
3 10
2.5 8.5
2.2 7.6
2.1 7.3
2.03 7.09
2.009 7.027
2.0005 7.0015
2.0000001 7.0000003

Observe that as the values of x get closer and closer to 2, the values of f (x) get closer and closer to 7. This
behavior can be shown no matter what set of values, or what direction, is taken in approaching 2. In symbols,

lim (1 + 3x) = 7.
x→2

Example 3.1. Investigate

lim (x2 + 1)
x→−1

by constructing table of values. Here, c = −1 and f (x) = x2 + 1.

We start again by approaching −1 from the left.

x f (x)
-1.5 3.25
-1.2 2.44
-1.01 2.0201
-1.0001 2.00020001

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3.1 Table of Values Page 3

Now approach −1 from the right.

x f (x)
-0.5 1.25
-0.8 1.64
-0.99 1.9801
-0.9999 1.99980001

The tables show that as x approaches -1, f (x) approaches 2. In symbols,

lim (x2 + 1) = 2.
x→−1

Example 3.2. Investigate lim |x| through a table of values.

x→0

Approaching 0 from the left and from the right, we get the following tables:

x |x|
-0.3 0.3
-0.01 0.01
-0.00009 0.00009
-0.00000001 0.00000001

x |x|
0.3 0.3
0.01 0.01
0.00009 0.00009
0.00000001 0.00000001

Hence, lim |x| = 0.

x→0

Example 3.3. Investigate

x2 − 5x + 4
lim
x→1 x−1
x2 − 5x + 4
by constructing table of values. Here, c = 1 and f (x) = . Take note that 1 is not in the domain of f ,
x−1
but this is not a problem. In evaluating a limit, remember that we only need to go very close to 1; we will not go
to 1 itself.

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3.1 Table of Values Page 4

We now approach 1 from the right.

x f (x)
1.5 -2.5
1.17 -2.83
1.003 -2.997
1.0001 -2.9999

Approach 1 from left.

x f (x)
0.5 -3.5
0.88 -3.12
0.996 -3.004
0.9999 -3.0001

The tables show that as x approaches 1, f (x) approaches -3. In symbols,

x2 − 5x + 4
lim = −3.
x→1 x−1

Example 3.4. Investigate through a table of values

lim f (x)
x→4

if (
x+1 if x < 4
f (x) =
(x − 4)2 + 3 if x ≥ 4.
This looks a bit different, but the logic and procedure are exactly the same. We still approach the constant 4 from
the left and from the right, but note that we should evaluate the appropriate corresponding functional expression.
In this case, when x approaches 4 from the left, the values taken should be substituted in f (x) = x + 1. Indeed,
this is the part of the function which accepts values less than 4. So,

x f (x)
3.7 4.7
3.85 4.85
3.995 4.995
3.99999 4.99999

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3.2 Looking at the graph Page 5

On the other hand, when x approaches 4 from the right, the values taken should be substituted in f (x) =
(x − 4)2 + 3. So,

x f (x)
4.3 3.09
4.1 3.01
4.001 3.000001
4.00001 3.0000000001

Observe that the values that f (x) approaches are not equal, namely, f (x) approaches 5 from the left while it
approaches 3 from the right. In such a case, we say that the limit of the given function does not exist (DNE). In
symbols,
lim f (x) DNE.
x→4

Remark 1. We need to emphasize an important fact. We do not say that lim f (x) “equals DNE”, nor do
x→4
we write lim f (x) = DNE, because “DNE” is not a value. In the previous example, “DNE” indicated that
x→4
the function moves in different directions as its variable approaches c from the left and from the right. In
1
other cases, the limit fails to exist because it is undefined, such as for lim which leads to division of 1 by
x→0 x
zero.

Remark 2. Have you noticed a pattern in the way we have been investigating a limit? We have been
specifying whether x will approach a value c from the left, through values less than c, or from the right,
through values greater than c. This direction may be specified in the limit notation, lim f (x) by adding
x→c
certain symbols.

• If x approaches c from the left, or through values less than c, then we write lim f (x).
x→c−

• If x approaches c from the right, or through values greater than c, then we write lim f (x).
x→c+

Furthermore, we say
lim f (x) = L
x→c

if and only if
lim f (x) = L and lim f (x) = L.
x→c− x→c+

In other words, for a limit L to exist, the limits from the left and from the right must both exist and be equal
to L. Therefore,
lim f (x) = DNE whenever lim f (x) 6= lim f (x).
x→c x→c− x→c+
These limits, lim f (x) and lim f (x) are referred to as one-sided limits.
x→c− x→c+

3.2 Looking at the graph

If one knows the graph of f (x), it will be easier to determine its limits as x approaches given values of c.

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3.2 Looking at the graph Page 6

Let us look at the examples again, one by one.

Recall Example 2.1 where f (x) = x2 + 1.

Use desmos.com to graph the function.

It can be seen from the graph that as values of x approach 1, the values of f (x) approach 2.

Recall Example 2.2 where f (x) = |x|.

Use desmos.com to graph the function.

It is clear that lim |x|, that is, the two sides of the graph both move downward to the origin (0,0) as x approaches
x→0
0.

Recall Example 2.4 where (

x+1 if x < 4
f (x) =
(x − 4)2 + 3 if x ≥ 4.
Use desmos.com to graph the function.
Again, we can see from the graph that f (x) has no limit as x approaches 4. The two separate parts of the function
move toward different y-levels (y = 5 from the left, y = 3 from the right) in the vicinity of c = 4.

So, in general, if we have the graph of a function, such as below, determining limits can be done much faster
and easier by inspection.

3.2.1 Test your knowledge!

Exercises 3.1.

1. Complete the table of values to investigate lim (x2 − 2x + 4).

x→1

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3.2 Looking at the graph Page 7

x−1
2. Complete the table of values to investigate lim .
x→0 x+1

3. Construct a table of values to investigate the following limits:

10 1
(a) lim (e) lim
x→3 x−2 x→1 x+1
10 
(b) lim
x→7 x − 2
x + 3
 if x < 1
2x + 1 (f) lim f (x) if f (x) = 2x if x = 1
(c) lim x→1 √

x→2 x − 3 5x − 1 if x > 1
x2 + 6
(d) lim 2
x→0 x − 2

4. Consider the function f (x) whose graph is shown below.

Determine the following:

(a) lim f (x)

x→−3
(b) lim f (x)
x→−1
(c) lim f (x)
x→−1
(d) lim f (x)
x→3

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3.2 Looking at the graph Page 8

(e) lim f (x)

x→5

5. Consider the function f (x) whose graph is shown below.

What can be said about the limit of f (x)?

(a) at c = 1 , 2, 3, and 4?
(b) at integer values of c?
(c) at c = 0.4, 2.3,4.7, and 5.5?
(d) at non-integer values of c?

Remark 3. It is hoped that by now you have observed that for polynomial and rational functions f ,
if c is in the domain of f , then to evaluate lim f (x) you just need to substitute the value of c for
x→c
every x in f (x).

However, this is not true for general functions. Can you give an example or point out an earlier
example of a case where c is in the domain of f , but lim 6= f (c)?
x→c

6. Without a table of values and without graphing f (x), give the values of the following limits and explain how
you arrived at your evaluation.

(a) lim (3x − 5)

x→−1

x2 − 9
(b) lim where c = 0, 1, 2
x→c x2 − 4x + 3
x2 − 9
(c) lim 2
x→3 x − 4x + 3

1
7. Consider the function f (x) = whose graph is shown below.
x

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3.2 Looking at the graph Page 9

What can be said about lim f (x)? Does it exist or not? Why?
x→0

Answer: The limit does not exist. From the graph itself, as x-values approach 0, the arrows move in
opposite directions. If tables of values are constructed, one for x-values approaching 0 through negative
values and another through positive values, it is easy to observe that the closer the x-values are to 0, the
more negatively and positively large the corresponding f (x)-values become.

8. Sketch one possible graph of a function f (x) defined on R that satisfies all the listed conditions.

(a) lim f (x) = 1 (d) f (1) = 2

x→0
(e) f (2) = 0
(b) lim f (x) DNE
x→1 (f) f (4) = 5
(c) lim f (x) = 0 (g) lim f (x) = 5 for all c > 4
x→2 x→c

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 Page 10

4 The Limit of a Function at c versus the Value of the Function at c

Critical to the study of limits is the understanding that the value of

lim f (x)
x→c

may be distinct from the value of the function at x = c, that is,f (c). As seen in previous examples, the limit may
be evaluated at values not included in the domain of f . Thus, it must be clear to a student of calculus that the
exclusion of a value from the domain of a function does not prohibit the evaluation of the limit of that function
at that excluded value, provided of course that f is defined at the points near c. In fact, these cases are actually
the more interesting ones to investigate and evaluate.

Furthermore, the awareness of this distinction will help the student understand the concept of continuity.

Let us again consider

lim (1 + 3x).
x→2

Recall that the values are

x f (x)
1 4
1.4 5.2
1.7 6.1
1.9 6.7
1.95 6.85
1.997 6.991
1.9999 6.9997
1.9999999 6.9999997

x f (x)
3 10
2.5 8.5
2.2 7.6
2.1 7.3
2.03 7.09
2.009 7.027
2.0005 7.0015
2.0000001 7.0000003

and we had concluded that lim (1 + 3x) = 7.

x→2
In comparison, f (2) = 7. So, in this example, lim f (x) and f (2) are equal. Notice that the same holds for the
x→2
next examples discussed:

lim f (x) f (c)

x→c
lim (x2 + 1) = 2 f (−1) = 2
x→−1
lim |x| = 0 f (0) = 0
x→0

4 THE LIMIT OF A FUNCTION AT C VERSUS THE VALUE OF THE FUNCTION

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This, however, is not always the case. Let us consider the function
(
|x| if x 6= 0
f (x) = .
2 if x = 0

In contrast to the second example above, the entries are now unequal:

lim f (x) f (c)

x→c
lim f (x) = 0 f (0) = 2
x→0

Does this in any way affect the existence of the limit? Not at all. This example shows that lim f (x) and f (c)
x→c
may be distinct.

4.0.1 Test your knowledge!

Exercises 4.1.

1. Consider the function f (x) whose graph is given below.

Based on the graph, fill in the table with the appropriate values.

c lim f (x) f (c)

x→c
-2
-1/2
0
1
3
4

2. Consider the function f (x) whose graph is given below.

4 THE LIMIT OF A FUNCTION AT C VERSUS THE VALUE OF THE FUNCTION

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Evaluate lim f (x) and f (c) and state whether they are equal or not.
x→c

(a) c = −3 (f) c = 2
(b) c = −2 (g) c = 2.3
(c) c = 0 (h) c = 3
(d) c = 0.5 (i) c = 4
(e) c = 1 (j) c = 6

5 Illustration of Limit Theorems

Limits can be determined through either a table of values or the graph of a function. One might ask: Must one
always construct a table or graph the function to determine a limit? Filling in a table of values sometimes requires
very tedious calculations. Likewise, a graph may be difficult to sketch. However, these should not be reasons for
a student to fail to determine a limit.

In this section, we will learn how to compute the limit of a function using Limit Theorems.

We are now ready to list down the basic theorems on limits. We will state eight(8) theorems. These will enable
us to directly evaluate limits, without need for a table or a graph. In the following statements, c is a constant,
and f and g are functions which may or may not have c in their domains.

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5.1 Limit Theorems Page 13

5.1 Limit Theorems

Limit of a Constant
Definition 5.1. The limit of a constant is itself. If k is any constant, then,

lim k = k.
x→c

Example 5.1.

1. lim 2 = 2
x→c

2. lim π = π
x→0

3. lim 900 = 900

x→π

Substitution
Definition 5.2. The limit of x as x approaches c is equal to c. This may be thought of as the substitution
law, because x is simply substituted by c.
lim x = c
x→c

Example 5.2.

1. lim x = 9
x→9

2. lim x = 0.0005
x→0.0005

3. lim x = −10
x→−10

For the remaining theorems, we will assume that the limits of f and g both exist as x approaches c and that
they are L and M , respectively. In other words,

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

x→c x→c

Constant Multiple Theorem

Definition 5.3. The limit of a multiple of a function is simply that multiple of the limit of the function.

lim [k · f (x)] = k · lim f (x) = k · L

x→c x→c

Example 5.3. If lim f (x) = 4, then

x→c

1. lim [8 · f (x)] = 8 · lim f (x) = 8 · 4 = 32

x→c x→c

2. lim [−11 · f (x)] = −11 · lim f (x) = −11 · 4 = −44

x→c x→c

3 3 3
3. lim [ · f (x)] = · lim f (x) = · 4 = 6
x→c 2 2 x→c 2

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5.1 Limit Theorems Page 14

Addition
Definition 5.4. The limit of a sum of functions is the sum of the limits of the individual functions.
Subtraction is also included in this law, that is, the limit of a difference of functions is the difference of their
limits.
lim [f (x) + g(x)] = lim f (x) + lim g(x) = L + M
x→c x→c x→c

lim [f (x) − g(x)] = lim f (x) − lim g(x) = L − M

x→c x→c x→c

Example 5.4. For example, if lim f (x) = 4 and lim g(x) = −5, then
x→c x→c

1. lim [f (x) + g(x)] = lim f (x) + lim g(x) = 4 + (−5) = −1

x→c x→c x→c

2. lim [f (x) − g(x)] = lim f (x) − lim g(x) = 4 − (−5) = 9

x→c x→c x→c

Multiplication

Definition 5.5. This is similar to the Addition Theorem, with multiplication replacing addition as the
operation involved. Thus, the limit of a product of functions is equal to the product of their limits.

lim [f (x) · g(x)] = lim f (x) · lim g(x) = L · M

x→c x→c x→c

Example 5.5. For example, if lim f (x) = 8 and lim g(x) = −1, then
x→c x→c

1. lim [f (x) · g(x)] = lim f (x) · lim g(x) = 8 · (−1) = −8

x→c x→c x→c

Remark 4. The Addition and Multiplication Theorems may be applied to sums, differences, and products
of more than two functions.

Remark 5. The Constant Multiple Theorem is a special case of the Multiplication Theorem. Indeed, in the
Multiplication Theorem, if the first function f (x) is replaced by a constant k, the result is the Constant
Multiple Theorem.

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5.1 Limit Theorems Page 15

Division
Definition 5.6. This says that the limit of a quotient of functions is equal to the quotient of the limits of
the individual functions, provided the denominator limit is not equal to 0.

f (x) lim f (x) L

lim = x→c = , provided M 6= 0
x→c g(x) lim g(x) M
x→c

Example 5.6.

1. For example, if lim f (x) = 3 and lim g(x) = −6, then

x→c x→c
f (x) lim f (x) 3 1
lim = x→c = =−
x→c g(x) lim g(x) (−6) 2
x→c

2. For example, if lim f (x) = 0 and lim g(x) = 1, then

x→c x→c
f (x) lim f (x) 0
lim = x→c = =0
x→c g(x) lim g(x) 1
x→c

3. For example, if lim f (x) = 3 and lim g(x) = 0, then

x→c x→c
f (x) lim f (x)
3
lim = x→c = = undefined. Thus, the limit does not exist.
x→c g(x) lim g(x) 0
x→c

Power
Definition 5.7. This theorem states that the limit of an integer power p of a function is just that power
of the limit of the function.
lim [f (x)]p = [lim f (x)]p = Lp
x→c x→c

Example 5.7.

1. If lim f (x) = 4, then

x→c
lim [f (x)]3 = [lim f (x)]3 = 43 = 64
x→c x→c

2. If lim f (x) = −2, then

x→c
lim [f (x)]5 = [lim f (x)]5 = (−2)5 = −32
x→c x→c

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5.1 Limit Theorems Page 16

Radical/Root

Definition 5.8. This theorem states that if n is a positive integer, the limit of the nth root of a function
is just the nth root of the limit of the function, provided the nth root of the limit is a real number. Thus,
it is important to keep in mind that if n is even, the limit of the function must be positive.
p q √
n
lim n f (x) = n lim f (x) = L
x→c x→c

Example 5.8.

1. If lim f (x) = 4, then

x→c p q √
lim f (x) = lim f (x) = 4 = 2
x→c x→c

2. If lim f (x) = −4, then

x→c p q √
lim f (x) = lim f (x) = −4.
x→c x→c

But this is not a real number, thus it is impossible to evaluate this limit.

5.1.1 Test your knowledge!

Exercises 5.1.

1. Assume the following:

3
lim f (x) = , lim g(x) = 12, lim h(x) = −3
x→c 4 x→c x→c

Compute the following limits:

(a) lim [−4 · f (x)] (g) lim [4 · f (x) + h(x)]

x→c x→c
p
(b) lim 12 · f (x) (h) lim [f (x) · g(x) · h(x)]
x→c x→c
(c) lim [g(x) − h(x)] p
x→c (i) lim −g(x) · h(x)
(d) lim [f (x) · g(x)] x→c
x→c  
g(x) + h(x) g(x)
(e) lim (j) lim
x→c (h(x))2
x→c f (x)
   
f (x) g(x)
(f) lim · g(x) (k) lim · f (x)
x→c h(x) x→c (h(x))2

2. Assume f (x) = x. Evaluate

(a) lim f (x) 1

x→4 (c) lim
x→0 (f (x))2

1 p
(b) lim (d) lim − f (x)
x→−1 f (x) x→3

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p
(e) lim 9 · f (x) (g) lim [(f (x))3 + (f (x)) + 2 · f (x)]
x→0.5 x→12
(f (x))2 − f (x)
(f) lim [(f (x))2 − f (x)] (h) lim
x→9 x→−1 5 · f (x)

6 Limits of Polynomial, Rational, and Radical Functions

In this lesson, we will show how these limit theorems are used in evaluating algebraic functions. Particularly, we
will illustrate how to use them to evaluate the limits of polynomial, rational and radical functions.

6.1 Limits of polynomial functions

We start with evaluating the limits of polynomial functions.

Example 6.1. Determine lim (2x + 1).

x→1

Solution 6.1.1. From the theorems above,

lim (2x + 1) = lim 2x + lim

x→1 x→1 x→1
= [2 · lim x] + 1
x→1
= 2(1) + 1
=2+1
= 3.

Example 6.2. Determine lim (2x3 − 4x2 + 1).

x→−1

Solution 6.2.1. From the theorems above,

lim (2x3 − 4x2 + 1) = lim 2x3 − lim 4x2 + lim 1

x→−1 x→−1 x→−1 x→−1
3 2
= 2 · lim x − 4 · lim x + 1
x→−1 x→−1
= 2 · ( lim x) − 4 · ( lim x)2 + 1
3
x→−1 x→−1
3 2
= 2(−1) − 4(−1) + 1
= −2 − 4 + 1
= −5

6.2 Limits of Rational Functions

We will now apply the limit theorems in evaluating rational functions. In evaluating the limits of such functions,
recall the Division Rule, and all the rules stated which have been useful in evaluating limits of polynomial functions,
such as the Addition and Product Rules.
1
Example 6.3. Evaluate lim .
x→1 x

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6.2 Limits of Rational Functions Page 18

Solution 6.3.1. First, note that lim x = 1. Since the limit of the denominator is non-zero, we can apply the
x→1
Division Rule. Thus,

1 lim 1
lim = x→1
x→1 x lim x
x→1
1
=
1
= 1.
x
Example 6.4. Evaluate lim .
x→2 x+1
Solution 6.4.1. We start by checking the limit of the polynomial function in the denominator.

lim (x + 1) = lim x + lim 1 = 2 + 1 = 3.

x→2 x→2 x→2

Since the limit of the denominator is not zero, it follows that

x lim x 2
x→2
lim = = .
x→2 x + 1 lim (x + 1) 3
x→2

(x − 3)(x2 − 2)
Example 6.5. Evaluate lim .
x→1 x2 + 1
Solution 6.5.1. First note that

lim (x2 + 1) = lim x2 + lim 1 = 1 + 1 = 2 6= 0.

x→1 x→1 x→1

Thus, using the theorem,

(x − 3)(x2 − 2) lim (x − 3)(x2 − 2)

x→1
lim =
x→1 x2 + 1 lim x2 + 1
x→1
lim (x − 3) · lim (x2 − 2)
x→1 x→1
=
 2  
lim x − lim 3 lim x2 − lim 2
x→1 x→1 x→1 x→1
=
2
(1 − 3)(12 − 2)
=
2
= 1.

Theorem 6.1. Let f be a polynomial of the form

f (x) = an xn + an−1 xn−1 + an−2 xn−2 + ... + a1 x + a0 .

If c is a real number, then

lim f (x) = f (c).
x→c

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6.3 Limit of Radical Functions Page 19

6.3 Limit of Radical Functions

We will now evaluate limits of radical functions using limit theorems.
√
Example 6.6. Evaluate lim x.
x→1

Solution 6.6.1. Note that lim x = 1 > 0. Therefore, by Radical/Root rule,

x→1
√ q √
lim x= lim x = 1 = 1.
x→1 x→1
√
3
Example 6.7. Evaluate lim x2 + 3x − 6.
x→−2

Solution 6.7.1. Since the index of the radical sign is odd, we do not have to worry that the limit of the radicand
is negative. Therefore, the Radical/Root Rule implies that
p
3
q √ √
lim x2 + 3x − 6 = 3 lim (x2 + 3x − 6) = 3 4 − 6 − 6 = 3 −8 = −2.
x→−2 x→−2

6.4 Infinite limits

Infinite Limits
Definition 6.1. The limit of f (x) as x approaches c is positive infinity, denoted by,

lim f (x) = +∞
x→c

if the value of f (x) increases without bound whenever the values of x gets closer and closer to c.

Similarly, the limit of f (x) as x approaches c is negative infinity, denoted by,

lim f (x) = −∞
x→c

if the value of f (x) decreases without bound whenever the values of x gets closer and closer to c.

Remark 6. Remember that ∞ is not a number. It holds no specific value. So, lim f (x) = +∞ or lim f (x) =
x→c x→c
−∞ describes the behavior of the function near x = c, but it does not exist as a real number.

6.4.1 Test your knowledge!

Exercises 6.1.

1. Evaluate the following limits.

√ √ √
(a) lim (1 + 3
w)(2 − w2 + 3w3 ) 2x − 6 − x
w→1 (d) lim
x→2 4 + x2
t2 − 1 
2z + z 2
3
(b) lim (e) lim
t→−2 t2 + 3t − 1
z→2 z2 + 4
4 − 3y 2 − y 3 x2 − x − 2
(c) lim (f) lim 3
y→−2 6 − y − y 2 z→0 x − 6x2 − 7x + 1

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6.4 Infinite limits Page 20

2. Recall the graph of y = csc x. From the behavior of the graph of the cosecant function, evaluate following
limits.

(a) lim csc x (c) lim csc x

x→0− x→π −
(b) lim csc x (d) lim csc x
x→0+ x→π +

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