LIMITS AND CONTINUITY

At the end of the chapter, the students are expected to:

a. Illustrate the limit of a function using a table of values and the graph of a function;
b. Identify and define the different Limit Theorems;
c. Determine the limits of a function by applying the different Limit Theorems;
d. Determine other techniques of evaluating limits of functions;
e. Find the limits of transcendental functions;
f. Illustrate continuity of a function; and
g. Illustrate different types of discontinuity.

The main focus of Calculus is to study how things change. Real-life situations can be modeled
using functions. Calculus provides a way to investigate the effects of these situations which deal with
change. This involves differentiation and integration. The main tool in studying the derivatives is the
concept of limits. Problems on determining long-term behavior make use of infinite limits. With this
concept, questions such as:

• What will happen to the level of population in the long run?

• If a certain medicine is injected into the bloodstream, how do we measure the concentration
of the medicine after an indefinite period of time?

can be investigated.

In Algebra, a fundamental problem was

finding the value of a function f(x) at x = c. In Calculus,
a basic problem is finding the values of a function f(x)
near x = c. The concept of limits is what distinguishes
Calculus from Algebra.
Lesson 1: LIMIT OF FUNCTIONS
Suppose we want to graph the function y = f(x) = 2x + 1. We use a table and include the
intercepts, if there are any, in the table.

1 1
x −6 −3 −1 − 0 1 4 5
2 2

y = 2x + 1 −11 −5 −1 0 1 2 3 9 11

2𝑥 2 −9𝑥−5
Consider the function 𝑦 = 𝑓(𝑥) = . If we factor the numerator, we obtain
𝑥−5
(2𝑥+1)(𝑥−5)
𝑦= = 2𝑥 + 1 if x ≠ 5.
𝑥−5

Some questions that need to be considered:

• What happens to the y-values as the x-values get closer to x ≠ 5?
• What is the difference between the graphs of the functions y = 2x + 1, where x ≠ 5 and the
previous example y = 2x + 1?

x 4 4.1 4.5 4.9 4.999 4.99999 5 5.00001 5.01 5.1 5.5 6

y = 2x + 1,
9 9.2 10 10.8 10.998 10.99998 undefined 11.00002 11.02 11.2 12 13
x≠ 5

As the x-values get very close to x=5 through numbers from the left of 5 and from the right of
5, the y-values approach 11. The points (5, 11) but the point itself is not defined.

(2𝑥+1)(𝑥−5)
The graph of 𝑦 = is the line y = 2x + 1 with a hole at the point (5, 11).
𝑥−5
 The hole at the point (5, 11) indicated a break in the line. We note that the functional value f(5)
is undefined but we can predict what happens to the y-values at x-values very near x = 5. We say that
2𝑥 2 −9𝑥−5
the limit of the function f(x) = as x approaches 5 is equal to 11. We denote this as
𝑥−5
2𝑥 2 − 9𝑥 − 5
lim = 11
𝑥→5 𝑥−5

Definition:
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

Example:
lim (1 + 3𝑥).
𝑥→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).
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:
√𝑥 2 +100 −10
Consider the function 𝑦 = 𝑓(𝑥) = . Suppose we want to estimate the value of
𝑥2
lim 𝑓(𝑥).
𝑥→0

x −0.01 −0.001 −0.0001 0 0.0001 0.001 0.01

1 1
y 0.0499999 = 0.05 0.04 undefined 0.0499999 = 0.05 0.04999
20 20

1
From the y-values, we can guess that lim 𝑓(𝑥) = = 0.05.
𝑥→0 20
Name: ___________________________________________________________ Score:
____________________
Year & Section: ___________________________________________________ Date:
______________________

Exercises:
Find the limit of the function below by supplying the table of values.

3
√𝑥 2 +4𝑥
1. lim
𝑥→0 𝑥

x -0.01 -0.001 -0.0001 0 0.0001 0.001 0.01

2𝑥 2 −1
2. lim
𝑥→1 𝑥−1

x -0.01 -0.001 -0.0001 0 0.0001 0.001 0.01

3. lim sin 2𝑥
𝑥→0

x -0.01 -0.001 -0.0001 0 0.0001 0.001 0.01

5𝑥
4. lim
𝑥→0 𝑥

x -0.01 -0.001 -0.0001 0 0.0001 0.001 0.01

√3𝑥 3
5. lim
𝑥→−2 𝑥+2

x -0.01 -0.001 -0.0001 0 0.0001 0.001 0.01

y
Lesson 2: THE LIMIT THEOREM

A. The Limit of a Constant Theorem

lim 𝑐 = 𝑐
𝑥→𝑘

Examples:
1. lim 3 = 3
𝑥→4
Remember:
2 2 The limit of a constant as x
2. lim1 =
𝑥→2 5 5 approaches to any constant is always equal to
the given constant.
3. lim 𝜋 = 𝜋
𝑥→3

4. lim 3.4 = 3.4

𝑥→1.2

5. lim 4 = 4
𝑥→−2

B. Limit of a Function x
lim 𝑥 = 𝑘
𝑥→𝑘

Examples:
1. lim 𝑥 = 1
𝑥→1
Remember:
2. lim 𝑥=
−1 The limit of the function x as x
−1 2
𝑥→ 2 approaches to any constant is always equal to
the constant.
3. lim 𝑥 = 3.25
𝑥→3.25

4. lim 𝑥 = 𝜋
𝑥→𝜋

5. lim 𝑥 = −5
𝑥→−5
C. Limit of a Constant and a Function f(x)
lim 𝑘 𝑓(𝑥) = 𝑘 ∗ lim 𝑓(𝑥)
𝑥→𝑘 𝑥→𝑘

Examples:
1. lim 3𝑥 = 3 lim 𝑥
𝑥→−2 𝑥→−2
Remember:
= 3(-2)
In evaluating the limit of a constant and a
= -6
function, the following steps can be followed:
1. Express the limit as a product of a constant and
2. lim1 − 6𝑥 = −6 lim1𝑥
𝑥→2 𝑥→2 the limit of a function x.
1 2. Find the limit of the function x based on the given
= -6( )
2 value for x.
= -3 3. Simplify the resulting number.

3. lim1 5𝑥 = 5 lim1 𝑥
𝑥→25 𝑥→25
1
=5( )
25
1
=
5

D. Limit of Sum and Difference of Functions

lim [𝑓(𝑥) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥)
𝑥→𝑘 𝑥→𝑘 𝑥→𝑘

Examples:
1. lim 𝑥 + 5 = lim 𝑥 + lim 5
𝑥→−2 𝑥→−2 𝑥→−2
= -2 + 5 Remember:
=3 In evaluating the limit of sum/difference of
functions, the following steps can be followed:
2. lim 4𝑥 − 2 = 4 lim 𝑥 − lim 2 1. Express the limit as a sum or difference of
𝑥→−5 𝑥→−5 𝑥→−5
= 4(-5) – 2 functions.
= -20 – 2 2. Apply the previously discussed limit theorems in
= -22 finding the limit of each term.
3. Simplify the resulting number.
3. lim1 − 3𝑥 − 4 = −3lim1𝑥 − lim1 4
𝑥→3 𝑥→3 𝑥→3
1
= -3( ) −4
3
= -1 – 4
= -5
E. Limit of Product of Functions
lim [𝑓(𝑥) ∙ 𝑔(𝑥)] = lim 𝑓(𝑥) ∙ lim 𝑔(𝑥)
𝑥→𝑘 𝑥→𝑘 𝑥→𝑘

Examples:
1. lim 3𝑥(5) = 3 lim 𝑥 ∙ lim 5
𝑥→1 𝑥→1 𝑥→1
= 3(1) (5)
= 3(5)
= 15

2. lim (2𝑥 + 1)(𝑥 + 5) = (2 lim 𝑥 + lim 1)( lim 𝑥 + lim 5)

𝑥→−2 𝑥→−2 𝑥→−2 𝑥→−2 𝑥→−2
= [2(-2) + 1] [-2 + (5)]
= (-4 + 1)(3)
= (-3)(3)
= -9

3. lim (𝑥 + 1)(−5) = (lim 𝑥 + lim 1)( lim − 5)

𝑥→−3 𝑥→−3 𝑥→−3 𝑥→−3
= (-3 + 1)(-5)
= (-2)(-5)
= 10

Remember:
In evaluating the limit of the products of functions,
the following steps can be followed:
1. Express the limit in expanded form by applying
the Limit Product Theorems.
2. Evaluate the limit of each function by applying the
previously discussed limit theorems.
3. Simplify the resulting number.

F. Limit of Quotient of Functions

𝑓(𝑥) 𝑥→𝑘lim 𝑓(𝑥)
lim = , 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑡ℎ𝑎𝑡 𝑔(𝑥) ≠ 0
𝑥→𝑘 𝑔(𝑥) lim 𝑔(𝑥)
𝑥→𝑘

Examples:
𝑥 lim 𝑥
1. lim = 𝑥→5
𝑥→5 2 lim 2
𝑥→5
5
=
2

3𝑥 3 lim𝑥
𝑥→−6
2. lim =
𝑥→−6 2 lim 2
𝑥→−6
3(−6) Remember:
=
2 In evaluating the limit of quotient of functions, the
−18
=
2
following steps can be followed:
1. Express the limit in expanded form by applying
the Limit of Quotients Theorems.
2. Evaluate the limit of each function by applying the
previously discussed limit theorems.
3. Simplify the resulting number.
 = -9

𝑥+5 lim 𝑥+ lim 5

𝑥→2 𝑥→2
3. lim =
𝑥→2 3𝑥−2 3 lim 𝑥− lim 2
𝑥→2 𝑥→2
2+5
=
3(2)−2
7
=
6−2
7
=
4

G. Limit of Power of Functions

lim [𝑓(𝑥)]𝑛 = [lim 𝑓(𝑥)]𝑛
𝑥→𝑘 𝑥→𝑘

Examples:
1. lim 𝑥 4 = [lim x]4
𝑥→2 𝑥→2
= 24
= 16

2. lim 𝑥 2 + 2𝑥 + 1 = [ lim 𝑥]2 + 2 lim 𝑥 + lim 1

𝑥→−9 𝑥→−9 𝑥→−9 𝑥→−9
= (-9)2 + 2(-9) + 1
= 81 – 18 + 1
= 64

3. lim 3𝑥 4 = 3 [lim 𝑥] 4
𝑥→−3 𝑥→−3
= 3(-3)4
= 3(81)
= 243

H. Limit of Root of Functions

𝑛
lim √𝑓(𝑥) = 𝑛√ lim (𝑓𝑥) , 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑡ℎ𝑎𝑡 lim 𝑓(𝑥) > 0. 𝑊ℎ𝑒𝑛 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛.
𝑥→𝑘 𝑥→𝑘

Examples:
1. lim √𝑥 = √ lim 𝑥
𝑥→8 𝑥→8

= √8
= √4(2)
= 2√2

3
2. lim 3𝑥 √3𝑥 = 3 lim 𝑥 ∙ 3√3 lim 𝑥
𝑥→27 𝑥→27 𝑥→27

= [3(27)] [ 3√3(27)]
3
= 81(3 √3)
3
= 243 √3
I. Substitution Theorem
Definition: If function f is a polynomial function or a rational function, then
lim 𝑓(𝑥) = lim 𝑓(𝑘)
𝑥→𝑘 𝑥→𝑘
provided that in case of rational functions, the value of the denominator at k is not equal to zero.

Examples:
1. Evaluate the lim 3𝑥 4 − 2𝑥 2 + 10𝑥 − 1
𝑥→2
Substitute all values of x by 2, then,
lim 3𝑥 4 − 2𝑥 2 + 10𝑥 − 1 = 3(2)4 − 2(2)2 + 10(2) − 1
𝑥→2
By simplifying, applying the PEMDAS rule, the answer will be 59.

2. Find the lim1 3𝑥 2 − 2𝑥 + 5

𝑥→3

1 2 1
lim 3𝑥 2 − 2𝑥 + 5 = 3 ( ) − 2 ( ) + 5
𝑥→13 3 3
1 2
= 3( ) − + 5
9 3
1−2+15
=
3
14
=
3

Remember:
In evaluating the limit of functions by Substitution Theorem, the following
steps can be followed:
1. Substitute all x’s by the given values.
2. Simplify by applying the PEMDAS Rule.
Name: ___________________________________________________________ Score:
____________________
Year & Section: ___________________________________________________ Date:
______________________

Exercises A:
Evaluate the following limits using the limit theorems except the Substitution method.

1. lim − 3
𝑥→−5

−3
2. lim
𝑥→5 5

3. lim 3𝜋
𝑥→0

3
4. lim √27
𝑥→−5

5. lim 𝑥
𝑥→−3

6. lim3 𝑥
𝑥→8

7. lim 24𝑥
𝑥→𝜋

8. lim 2𝑥
𝑥→−1

9. lim 8𝑥
𝑥→8

10. lim 𝑥
𝑥→−2

11. lim (2𝑥 − 15) 12. lim1 (−8𝑥 − 4)

𝑥→−1 𝑥→2

13. lim (𝑥 − 2) 14. lim1(6𝑥 + 12)

𝑥→−2 𝑥→−2
Exercises B.
Evaluate the following limits by applying the different limit theorems discussed.

1. lim 𝑥(2𝑥) 2. lim1(−6𝑥)(4𝑥)

𝑥→−3 𝑥→3

3. lim (13𝑥 − 2)(𝑥 + 1) 4. lim (3𝑥 − 2)(𝑥 + 7)

𝑥→7 𝑥→5

𝑥
5. lim ( ) 6. lim (𝑥 5 + 1)
𝑥→−7 −2 𝑥→5

4𝑥−1 𝑥−3
7. lim ( ) 8. lim ( )
𝑥→−3 𝑥−3 𝑥→−1 2𝑥+4

𝑥+√2
9. lim 10. lim 𝑥 3
𝑥→2 2 𝑥→4
Exercise C.
Evaluate the following limits by applying the different limit theorems discussed.

1. lim 𝑥 2 − 1 2. lim3 𝑥 4
𝑥→12 𝑥→5

3. lim1 2𝑥 3 + 4 4. lim 𝑥 3 + 2𝑥 − 10
𝑥→4 𝑥→−3

5. lim 𝑥 3 + 𝑥 2 − 𝑥 6. lim √𝑥 (2𝑥)

𝑥→−2 𝑥→20

√𝑥 + 5
7. lim 3𝑥 √𝑥 8. lim
𝑥→5 𝑥→15 2

3
9. lim (2𝑥 − 5)( √𝑥 + 5) 10. lim √𝑥 3 + 2𝑥 − 10
𝑥→−1 𝑥→2
Exercise D.
Evaluate the following limits by applying the Substitution Theorem.

1. lim 𝑥 − 2 2. lim √4𝑥 2 − 2𝑥 + 10

𝑥→2 𝑥→2

6𝑥+5
3. lim 3𝑥 4 − 2𝑥 3 + 3𝑥 2 − 10 4. lim
𝑥→5 𝑥→−3 3𝑥+10

2𝑥 2 +3𝑥+1 √7𝑥+2
5. lim 6. lim
𝑥→−1 𝑥−1 𝑥→2 𝑥

4𝑥 2 −3𝑥 𝑥+5
7. lim 8. lim
𝑥→−1 7𝑥−1 𝑥→0 𝑥−1

𝑥 4 −2𝑥+4
9. lim 10. lim2 (3𝑥 + 1)
𝑥→−1 √𝑥+5 𝑥→5
Other Techniques in Evaluating Limits of Functions

There are instances that applying the Substitution Theorem is not applicable. That is, when the
limit becomes undefined upon substituting the given value of x. In this case, other techniques such as
factoring method or conjugate method can be used.

J. By Factoring

Example:
𝑥 2 +3𝑥−10
1. Evaluate the lim .
𝑥→−5 𝑥+5

Using the Substitution Theorem,

𝑥 2 +3𝑥−10 (−5)2 +3(−5)−10

lim =
𝑥→−5 𝑥+5 −5+5
25−25−10
=
0
= undefined

By factoring,
𝑥 2 +3𝑥−10 (𝑥+5)(𝑥−2)
lim will become lim
𝑥→−5 𝑥+5 𝑥→−5 𝑥+5

Cancelling the common factor x+5, the given limit will now become
lim 𝑥 − 2.
𝑥→−5

By Substitution Theorem, the limit of the given function is -7.

Remember:
Factoring Method is used when:
a. The limit of the given rational function is undefined.
b. Either the numerator or denominator of the rational function is factorable.
In evaluating the limit of functions by Factoring Method, the following steps can be followed:
1. Factor either the numerator or denominator of the given rational function.
2. Cancel the common factor.
3. Apply Substitution Theorem then simplify.
K. By Conjugation
Conjugation is a process of simplifying rational expressions which contains radicals on the
denominator.
For instance, we are given the expression below and we are asked to simplify,
4
5 + √3

Steps for conjugation:

4 5−√3 20−4√3 20−4√3
∙ = =
5+√3 5−√3 25−3 22

Examples:
𝑥−25
1. Find lim .
𝑥→25 √𝑥−5

By conjugation,
𝑥−25 √𝑥+5 (√𝑥+5)(𝑥−25)
∙ = = √𝑥 + 5
√𝑥−5 √𝑥+5 𝑥−25

Getting the limit of √𝑥 + 5,

lim √𝑥 + 5 = √25 + 5
𝑥→25
=5+5
= 10
Therefore, the limit of the function as x approaches to 25 is 10.

11−√𝑥
2. Find lim .
𝑥→121 121−𝑥

By conjugation,
11−√𝑥 11+√𝑥 121−𝑥 1
∙ = (121−𝑥)+(11+ =
121−𝑥 11+√𝑥 √𝑥) 11+√𝑥

1
Taking the limit of ,
11+√𝑥
1 1
lim = 11+√121
𝑥→121 11+√𝑥
1
=
11+11
1
=
22

𝟏
Therefore, the limit of the function as x approaches to 121 is .
𝟐𝟐
Remember:
Conjugation Method is used when:
a. The limit of the given rational function is undefined; and
b. Either the numerator or the denominator of the rational function is composed of
radicals.
In evaluating the limit of functions by Conjugation Method, the following steps can be applied:
1. Get the conjugate of the radical.
2. Multiply the conjugate to the numerator and denominator of the given rational function.
3. Cancel the common factors existing in the numerator and denominator.
4. Apply the Substitution Theorem in order to get the limit of the result in step 3.
5. Simplify.
Name: ___________________________________________________________ Score:
____________________
Year & Section: ___________________________________________________ Date:
______________________

Exercises A.
Evaluate the following limits by applying the factoring method.

𝑥 2 +8𝑥+16 𝑥+5
1. lim 2. lim
𝑥→1 𝑥−1 𝑥→−5 𝑥 3 +125

𝑥 3 −8 𝑥+6
3. lim 4. lim
𝑥→2 𝑥−2 𝑥→−6 𝑥 2 +13𝑥+42

𝑥 2 −4𝑥+12 𝑥−3
5. lim 6. lim
𝑥→−2 𝑥+2 𝑥→3 𝑥 2 −18𝑥+45

𝑥 2 +4𝑥−32 𝑥 2 −64
7. lim 8. lim
𝑥→−8 𝑥+8 𝑥→−8 𝑥+8

𝑥 2 −22𝑥+121 𝑥−3
9. lim 10. lim
𝑥→11 𝑥−11 𝑥→−3 𝑥 2 −9
Exercise B.
Evaluate the following limits by applying the conjugate method.

𝑥−1 4
1. lim ( ) 2. lim ( )
𝑥→1 √𝑥−1 𝑥→5 √𝑥

√𝑥−1 𝑥−64
3. lim ( ) 4. lim ( )
𝑥→1 𝑥+1 𝑥→64 −√𝑥+8

𝑥−9
5. lim ( ) 6. lim (√4𝑥 2 + 2𝑥 + 10)
𝑥→9 √𝑥−3 𝑥→−2

−2+√𝑥 −3
7. lim ( ) 8. lim ( )
𝑥→4 𝑥+4 𝑥→8 √5𝑥

√𝑥−10
9. lim √𝑥 2 + 2𝑥 + 1 10. lim ( )
𝑥→1 𝑥→−3 𝑥−100
 Some functions have also vertical or horizontal asymptotes. When the limit of these functions
approaches to its asymptotes, the limit is infinite (or denoted by ∞).

L. Infinite Limit Theorem 1

1 1 −∞ 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
a. lim+ = +∞ b. lim− ={ }
𝑥→0 𝑥𝑛 𝑥→0 𝑥𝑛 +∞ 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛

Examples:
1
1. Find lim+
𝑥→0 𝑥4

Since the numerator = 1 and x approaches 0 from the right, Infinite Limit Theorem 1a
1
is applied. Therefore, lim+ = +∞.
𝑥→0 𝑥4

1
2. Evaluate: lim−
𝑥→0 2𝑥 3

Since the numerator = 1and the denominator of the given expression has a power of 3
which is an odd number, Infinite Limit Theorem 1b is applied.
1
Therefore, lim− = −∞
𝑥→0 2𝑥 3

M. Infinite Limit Theorem 2

𝑓(𝑥)
Let c be any real number. Suppose in lim , lim 𝑔(𝑥) = 0 and lim 𝑓(𝑥) = 𝑘, where k is a
𝑥→𝑐 𝑔(𝑥) 𝑥→𝑐 𝑥→𝑐
real number and k ≠ 0.
𝑓(𝑥)
a. If k > 0 and g(x) → 0 through positive values, then lim = +∞.
𝑥→𝑐 𝑔(𝑥)
𝑓(𝑥)
b. If k > 0 and g(x) → 0 through negative values, then lim = −∞.
𝑥→𝑐 𝑔(𝑥)
𝑓(𝑥)
c. If k < 0 and g(x) → 0 through positive values, then lim = −∞.
𝑥→𝑐 𝑔(𝑥)
𝑓(𝑥)
d. If k < 0 and g(x) → 0 through negative values, then lim = +∞.
𝑥→𝑐 𝑔(𝑥)

The Infinite Limit Theorem 2 also holds for limits where x → c+ and x → c-.

Examples:
2𝑥−1
1. Evaluate lim +
𝑥→−2 𝑥+2

Through direct substitution, we have lim + (2𝑥 − 1) = −5. From the denominator,
𝑥→−2
the lim + (𝑥 + 2) = 0 through positive values. Therefore, by the Infinite Limit
𝑥→−2
2𝑥−1
Theorem 2c, we have, lim + = −∞.
𝑥→−2 𝑥+2
𝑥3
2. Evaluate lim−
𝑥→3 (𝑥−3)2

Since lim− 𝑥 3 = 33 = 27, and lim−(𝑥 − 3)2 = 0 through positive values, then by the
𝑥→3 𝑥→3
𝑥3
Infinite Limit Theorem 2a, we have lim− = +∞.
𝑥→3 (𝑥−3)2
 5𝑥 2
3. Find the limit: lim+
𝑥→4 16−𝑥 2

We have lim+ 5𝑥 2 = 80. From the denominator, we have lim+ 16 − 𝑥 2 = 0 through

𝑥→4 𝑥→4
5𝑥 2
negative values. Thus, lim+ 2 = −∞.
𝑥→4 16−𝑥

N. Limit at Infinity Theorem 1

Let r be any positive integer. Then:
1
a. lim =0
𝑥→+∞ 𝑥 𝑟

1
b. lim =0
𝑥→−∞ 𝑥 𝑟

O. Limit at Infinity Theorem 2

Let n be a positive real number and k, any real number except 0. Then:
𝑘
a. lim =0
𝑥→+∞ 𝑥 𝑛

𝑘
b. lim =0
𝑥→−∞ 𝑥 𝑛

+∞ 𝑖𝑓 𝑘 > 0
c. lim 𝑘 ∙ 𝑥 𝑛 = {
𝑥→+∞ −∞ 𝑖𝑓 𝑘 < 0

+∞ 𝑖𝑓 𝑘 > 0 𝑎𝑛𝑑 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑜𝑟

𝑖𝑓 𝑘 < 0 𝑎𝑛𝑑 𝑛 𝑖𝑠 𝑜𝑑𝑑.
d. lim 𝑘 ∙ 𝑥 𝑛 = {
𝑥→−∞ −∞ 𝑖𝑓 𝑘 > 0 𝑎𝑛𝑑 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑜𝑟
𝑖𝑓 𝑘 < 0 𝑎𝑛𝑑 𝑛 𝑖𝑠 𝑜𝑑𝑑.

P. Limit at Infinity Theorem 3

𝑓(𝑥)
Let h(x) = , where f(x) and g(x) are polynomials.
𝑔(𝑥)

a. If the degree of f(x) is less than the degree of g(x), then

𝑓(𝑥) 𝑓(𝑥)
lim =0 lim =0
𝑥→+∞ 𝑔(𝑥) 𝑥→−∞ 𝑔(𝑥)

b. If the degree of f(x) is equal to the degree of g(x), and a is the leading coefficient of f(x) and
b is the leading coefficient of g(x), then
𝑓(𝑥) 𝑎 𝑓(𝑥) 𝑎
lim = lim =
𝑥→+∞ 𝑔(𝑥) 𝑏 𝑥→−∞ 𝑔(𝑥) 𝑏

c. If the degree of f(x) is greater than the degree of g(x), then

𝑓(𝑥)
lim = +∞ 𝑜𝑟 − ∞
𝑥→+∞ 𝑔(𝑥)
 𝑓(𝑥)
lim = +∞ 𝑜𝑟 − ∞
𝑥→−∞ 𝑔(𝑥)

Examples:
Find the limit of the following:
1
1. lim
𝑥→+∞ 5
𝑥
We apply Limit at Infinity Theorem 1a. Thus,
1
lim = 0.
𝑥→+∞ 𝑥 5

1
2. lim
𝑥→−∞ 2𝑥 8
We apply Limit at Infinity Theorem 1b. Thus,
1
lim = 0.
𝑥→−∞ 2𝑥 8

3
3. lim
𝑥→+∞ 𝑥 4
We apply Limit at Infinity Theorem 2a, to get
3
lim = 0.
𝑥→+∞ 𝑥 4

−12
4. lim
𝑥→−∞ 7𝑥 3
We apply Limit at Infinity Theorem 2b, we obtain
−12
lim = 0.
𝑥→−∞ 7𝑥 3

5. lim 5𝑥 4
𝑥→+∞
We apply Limit at Infinity Theorem 2c, we have
lim 5𝑥 4 = +∞
𝑥→+∞

6. lim 11𝑥 5
𝑥→−∞
We apply Limit at Infinity Theorem 2d, we get
lim 5𝑥 4 = −∞
𝑥→−∞

3𝑥 4 −5𝑥 3 +2𝑥
7. lim
𝑥→+∞ 2𝑥 5 +4𝑥 2 −7
Here, we notice that the highest degree is at the denominator [or g(x)], so we apply
Limit
at Infinity Theorem 3a. Hence,
3𝑥 4 −5𝑥 3 +2𝑥
lim = 0.
𝑥→+∞ 2𝑥 5 +4𝑥 2 −7

5𝑥 3 +16𝑥 2 −10
8. lim
𝑥→−∞ 12𝑥 3 +𝑥+28
Factor the highest-degree term out of the numerator and the denominator. Then, apply
the necessary Limit Theorems and Limit at Infinity Theorem 2.
 16 10
5𝑥 3 +16𝑥 2 −10 5𝑥 3 (1+5𝑥− 3 )
5𝑥
lim = lim 1 28
𝑥→−∞ 12𝑥 3 +𝑥+28 𝑥→−∞ 12𝑥 3 (1+ + )
12𝑥2 12𝑥3

16 2
5𝑥 3 (1+5𝑥− 3)
𝑥
= lim 1 7
𝑥→−∞ 12𝑥 3 (1+ + )
12𝑥2 3𝑥3
16 2 1 7
Since, lim = 0, lim = 0, lim = 0, and lim = 0, then:
𝑥→−∞ 5𝑥 𝑥→−∞ 𝑥 3 𝑥→−∞ 12𝑥 2 𝑥→−∞ 3𝑥 3

5𝑥 3 +16𝑥 2 −10 5𝑥 3 (1+0−0)

lim = lim
𝑥→−∞ 12𝑥 3 +𝑥+28 𝑥→−∞ 12𝑥 3 (1+0+0)
5
= lim
𝑥→−∞ 12
5
=
12

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 the previous examples,
the limit may be evaluated at values not included in the domain of f. Thus, it must be clear 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 that f is defined at the points near c.

Examples:
1. Evaluate lim (1 + 3𝑥)
𝑥→2

In comparison, the limit of the given function equals 7 as x approaches near 2. On the
other hand, evaluating the function at f(2) gives us 7 also.
However, this case is not always true.

𝑥 2 −5𝑥+4
2. Investigate lim
𝑥→1 𝑥−1

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

In this case, the limit of the function when x approaches 1 is -3. However, evaluating
the function at f(1) will give us 0 in the denominator.
Name: ___________________________________________________________ Score:
____________________
Year & Section: ___________________________________________________ Date:
______________________

Exercise A.
Evaluate the following limits applying the Infinite Limit Theorem.

9 2𝑡
1. lim− 2. lim−
𝑥→3 (𝑥−3)5 𝑡→−6 6+𝑡

𝑧+3 𝑔+7
3. lim− 4. lim+
𝑧→−1 (𝑧+1)2 𝑔→2 𝑥 2 −4

7𝑥 𝑥 2 −5𝑥+6
5. lim
10−
6. lim−
𝑥→ 3 (10−3𝑥)4 𝑥→−4 𝑥 3 +𝑥 2 −12𝑥

−2𝑥+1 𝑥 2 −4
7. lim+ 8. lim+
𝑥→3 𝑥−3 𝑥→−2 𝑥+4

1 1
9. lim− 10. lim+
𝑥→0 𝑥3 𝑥→0 𝑥2
Exercise B.
Evaluate the following limits applying the Limit of Infinity Theorem.

𝑥+2 3𝑥 2
1. lim 6. lim
𝑥→∞ 𝑥 2 +𝑥+1 𝑥→∞ 3𝑥 2 −2

2𝑥 2 3𝑥 2
2. lim 7. lim
𝑥→∞ 𝑥 2 −4 𝑥→∞ 4𝑥+4

2𝑥 3
3. lim (𝑥 3 − 4𝑥 2 + 5) 8. lim
𝑥→∞ 𝑥→∞ 3𝑥 2 −4

𝑥3 𝑥+1
4. lim 9. lim
𝑥→∞ 4𝑥 2 +3 𝑥→∞ 2𝑥 2 +2𝑥+1

√2𝑥 2 +3 √2𝑥 2 +3
5. lim 10. lim
𝑥→∞ 2𝑥+3 𝑥→∞ 4𝑥+2
Lesson 3: LIMIT OF SOME
TRANSCENDENTAL FUNCTIONS
EXPONENTIAL FUNCTIONS
First, we consider the natural exponential function f(x) = ex, where e is called the Euler
number, and has value 2.718281…

Example:
1. Evaluate the lim 𝑒 𝑥
𝑥→0

We construct the table of values for f(x) = ex for both sides of the limit.
x f(x)

-1 0.36787944117
-0.5 0.60653065971 x f(x)
-0.1 0.90483741803
1 2.71828182846
-0.01 0.99004983374
0.5 1.6487212707
-0.001 0.99900049983
0.1 1.10517091808
-0.0001 0.999900049983
0.01 1.01005016708
-0.00001 0.99999000005
0.001 1.00100050017
0.0001 1.000100005
Combining the two one-sided limits 0.00001 1.00001000005 allows
us to conclude that
lim 𝑒 𝑥 = 1
𝑥→0

Limit Laws on Exponential Function

1. If b > 1, then
lim 𝑏 𝑥 = 0 and lim 𝑏 𝑥 = +∞
𝑥→−∞ 𝑥→+∞

2. If 0 < b < 1, then

lim 𝑏 𝑥 = +∞ and lim 𝑏 𝑥 = 0
𝑥→−∞ 𝑥→+∞

3. Since e > 1, then

lim 𝑒 𝑥 = 0 and lim 𝑒 𝑥 = +∞
𝑥→−∞ 𝑥→+∞

Limit Definition for e

𝑒 𝑥 −1 1 𝑥
1. lim =1 3. lim (1 + ) = 𝑒
𝑥→0 𝑥 𝑥→+∞ 𝑥

1
2. lim (1 + 𝑥)𝑥 = 𝑒
𝑥→0

LOGARITHMIC FUNCTIONS
Now, consider the natural logarithmic function f(x) = ln x. Recall that ln x = logex.
Example 1. Evaluate lim ln x.
𝑥→1

We construct the table of values for f(x) = ln x for both sides of the limit.
x f(x) x f(x)

0.1 -2.30258509299 2 0.69314718056

0.5 -0.69314718056 1.5 0.4054651081
0.9 -0.10536051565 1.1 0.09531011798
0.99 -0.01005033585 1.01 0.00995033085
0.999 -0.00100050033 1.001 0.00099950033
0.9999 -0.000100005 1.0001 0.000099995
0.99999 -0.00001000005 1.00001 0.00000999995

Intuitively, lim ln x = 0. As the values of x gets closer and closer to 1, the values of f(x) get
𝑥→1
closer and closer to 0. Thus,
lim ln x = 0
𝑥→1

We now consider the common logarithmic function f(x) = log10 x. Recall that f(x) = log10x is
equal to log x.

Example 2. Evaluate lim log x.

𝑥→1

Constructing the table of values for both sides of the limit.

x f(x) x f(x)

0.1 -1 2 0.30102999566
0.5 -0.30102999566 1.5 0.17609125905
0.9 -0.04575749056 1.1 0.04139268515
0.99 -0.0043648054 1.01 0.00432137378
0.999 -0.00043451177 1.001 0.00043407747
0.9999 -0.00004343161 1.0001 0.00004342727
0.99999 -0.00000434296 1.00001 0.00000434292

As the values of x gets closer and closer to 1, the values of f(x) get closer and closer to
0.
In symbol:
lim log x = 0
𝑥→1
Remember:
1. lim log x = −∞ 3. lim ln x = +∞
𝑥→0 𝑥→+∞

2. lim ln x = −∞
𝑥→0

Limit Laws on Logarithmic Function

1. If b > 1, then
lim+ log 𝑏 𝑥 = −∞ and lim log 𝑏 𝑥 = +∞
𝑥→0 𝑥→+∞

2. If 0 < b 1, then
lim+ log 𝑏 𝑥 = +∞ and lim log 𝑏 𝑥 = −∞
𝑥→0 𝑥→+∞

TRIGONOMETRIC FUNCTIONS

Limit Theorems on Trigonometric Functions Special Trigonometric Limit Theorems

sin 𝜃
1. lim sin 𝑥 = sin 𝑐 7. lim =1
𝑥→𝑐 𝑥→𝜃 𝜃

𝜃
2. lim cos 𝑥 = cos 𝑐 8. lim =1
𝑥→𝑐 𝑥→𝜃 sin 𝜃

1−cos 𝜃
3. lim tan 𝑥 = tan 𝑐 9. lim =0
𝑥→𝑐 𝑥→𝜃 𝜃

4. lim cot 𝑥 = cot 𝑐

𝑥→𝑐

5. lim sec 𝑥 = sec 𝑐

𝑥→𝑐

6. lim csc 𝑥 = csc 𝑐

𝑥→𝑐

We now verify the three other Limit Theorems using a table of values where 𝜃 approaches 0
through positive and negative values.
𝜃
−0.01 −0.10 −0.001 0 0.10 0.01 0.001
(in radians)

sin 𝜃
0.9983 0.99998 0.9999998 undefined 0.9983 0.99998 0.9999998
𝜃
𝜃
1.0017 1.000017 1.00000017 undefined 1.0017 1.000017 1.00000017
sin 𝜃
1 − cos 𝜃
−0.0499 −0.00499 −0.000499 undefined 0.0499 0.00499 0.000499
𝜃

We can also find lim sin 𝑥 by using the graph of the sine function. Consider the graph of
𝑥→0
f(x) = sin x.
 Also, using the same graph, we have the following:
(a) lim𝜋 sin 𝑥 = 1 (c) lim𝜋 sin 𝑥 = −1
𝑥→ 2 𝑥→− 2

(b) lim sin 𝑥 = 0 (d) lim sin 𝑥 = 0

𝑥→𝜋 𝑥→−𝜋

Example 1. Evaluate lim𝜋 𝑥 sin 𝑥

𝑥→ 2
𝜋 𝜋
lim𝜋 𝑥 sin 𝑥 = ∙ sin
𝑥→ 2 2 2
𝜋
= ∙1
2
𝜋
=
2

tan 3𝑥
Example 2. Find the lim𝜋 2
𝑥→ 4 cot 2𝑥−sec 𝑥
3𝜋
tan 3𝑥 tan 4
lim𝜋 = 𝜋 𝜋
𝑥→ 4 cot 2𝑥−sec2 𝑥 cot 2( 4 )−sec2 ( 4 )

−1 1
= =
0−(√2)2 2

sin 5𝑥
Example 3. Find lim
𝑥→0 𝑥

To apply the Trigonometric Theorem 7, let’s introduce an allowable operation to have

5
5x in the denominator, so we multiply the expression with .
5
sin 5𝑥 5 sin 5𝑥
lim ∙ = ∙5
𝑥→0 𝑥 5 5𝑥
sin 5𝑥
lim ∙ lim 5
𝑥→0 5𝑥 𝑥→0
=1∙5
=5

1−cos 3𝑥
Example 4. Evaluate lim
𝑥→0 5𝑥

To apply Trigonometric Limit 9, we have to multiply and divide by 3.

1−cos 3𝑥 3
lim ∙
𝑥→0 5𝑥 3

1−cos 3𝑥 3
= lim ∙ (Interchange 3 and 5)
𝑥→0 3𝑥 5
 3
=0∙ =0
5
Name: ___________________________________________________________ Score:
____________________
Year & Section: ___________________________________________________ Date:
______________________

Exercise A.
Find the limits of the following functions.

2 1
sin3𝑥 sin2𝑥
1. lim 2. lim
𝑥→0 𝑥 𝑥→0 sin 4𝑥

tan 4𝑥
3. lim𝜋 4. lim 𝑒 5𝑥
𝑥→ 3 csc 2𝑥 𝑥→2

cos2 𝑥
5. lim (𝜃 tan 𝜃) 6. lim𝜋 ( )
𝜃→0 𝑥→ 2 1−sin 𝑥

1 𝑥 3𝑥 𝑒 𝑥 −1
7. lim (1 + ) (log 𝑥) 8. lim ( )( )
𝑥→+∞ 𝑥 𝑥→0 sin 𝑥 𝑥

1 𝑥 ln 𝑥
9. lim ( ) 10. lim
𝑥→+∞ 3 𝑥→+∞ √2
Lesson 4: CONTINUITY OF FUNCTIONS
What does “continuity at a point” mean? Intuitively, this means that in drawing the graph of a
function, the point in question will be traversed. We start by graphically illustrating what it means to be
continuity at a point.

Example 1. Consider the graph of f(x) = 3x – 1 below.

Question: Is the function continuous at x = 1?

To check if the function is continuous at x = 1, use the given graph. Note that one is able to
trace the graph from the left side of the number x = 1 going to the right side of x = 1, without lifting
one’s pen. This is the case here. Hence, we can say that the function is continuous at x = 1.

3𝑥 2 −4𝑥+1
Example 2. Now consider the graph of 𝑓(𝑥) = at the
𝑥−1
right.

Question: Is the function continuous at x = 1?

Tracing the graph from the left of x = 1 going to right of x
= 1, one finds that s/he must lift her/his pen briefly upon reaching
x = 1, creating a hole in the graph. Thus, the function is
discontinuous at x = 1.
 1
Example 3. Consider the graph of the function ℎ(𝑥) = .
𝑥

Is the function continuous at x = 0?

If we trace the graph from the left of x = 0 going to right of x = 0, we have to lift our pen since
at the left of x = 0, the function values will go downward indefinitely, while at the right of x = 0, the
function values will go to upward indefinitely. In other words,
1 1
lim+ = ∞ and lim− = −∞
𝑥→0 𝑥 𝑥→0 𝑥

Thus, the function is discontinuous at x = 0.

THREE CONDITIONS OF CONTINUITY

The function f(x) is continuous at a number c if and only if the following three conditions hold:
i. lim 𝑓(𝑥) exists;
𝑥→𝑐

ii. f(c) exists; and

iii. lim 𝑓(𝑥) = f(c)
𝑥→𝑐

The function f(x) is discontinuous at x = c if at least one of the three conditions is not satisfied.

Example: Determine if 𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 2 is continuous or not at x = 1.

(a) If x = 1, then f(1) = 0
(b) lim 𝑓(𝑥) = lim (𝑥 3 + 𝑥 2 − 2) = 13 + 12 − 2 = 0
𝑥→1 𝑥→1
(c) f(1) = 0 = lim 𝑓(𝑥)
𝑥→1

Therefore, f is continuous at x = 1.

Seatwork: Determine if the following functions are continuous at the given value of x.

1. 𝑓(𝑥) = 3𝑥 2 + 2𝑥 + 1 at x = -2

2. 𝑓(𝑥) = 9𝑥 2 − 1 at x = 1
 1
3. 𝑓(𝑥) = at x = 2
𝑥−2

𝑥−1
4. ℎ(𝑥) = at x = 1
𝑥 2 −1

𝑥+1
5. ℎ(𝑥) = at x = 1
𝑥 2 −1

ONE-SIDED CONTINUITY

(a) A function f is said to be continuous from the left at x = c if

𝑓(𝑐) = lim− 𝑓(𝑥).
𝑥→𝑐

(b) A function f is said to be continuous from the right at x = c if

𝑓(𝑐) = lim+ 𝑓(𝑥).
𝑥→𝑐

CONTINUITY OF POLYNOMIAL, ABSOLUTE VALUE, RATIONAL AND SQUARE ROOT

FUNCTIONS

(a) Polynomial functions are continuous everywhere.

(b) The absolute value function f(x) = |x| is continuous everywhere.
(c) Rational functions are continuous on their respective domains.
(d) The square root function f(x) = √𝑥 is continuous on [0, ∞).

CONTINUITY ON AN INTERVAL
Most of the functions that we shall be dealing with are continuous at all points on an interval
on the x-axis. If a function is continuous at all points in an open interval (a, b), then the function is
continuous on that interval.

Before continuing, let us review the interval set of values.

Case 1. Open Interval

The set of values of the variable x consists of all real numbers that lie between two fixed
numbers a and b, where a and b are not included.

{x | 𝑎 < 𝑥 < 𝑏}
(a, b)

Case 2. Closed Interval

The set of values of real numbers between a and b including the endpoints.
 {x | 𝑎 ≤ 𝑥 ≤ 𝑏}
[a, b]

Case 3. Half-Open Interval

The set of values of x including the endpoint a or b but not both endpoints.

{x | 𝑎 ≤ 𝑥 < 𝑏}
[a, b)

{x | 𝑎 < 𝑥 ≤ 𝑏}
(a, b]

Example. Consider the graph of function f given below.

 Using the graph, determine if the function f is continuous on the following intervals:
a. (-1, 1) b. (−∞, 0) c. (0, +∞)

Solution:
a. We can trace the graph from the right side of x = −1 to the left side of x = 1 without lifting the
pen we are using. Hence, we can say that the function f is continuous on the interval (−1, 1).
b. If we trace the graph from any negatively large number up to the left side of 0, we will not lift
our pen and so, f is continuous on ((−∞, 0).
c. For the interval (0,+∞), we trace the graph from the right side of 0 to any large number, and
find that we will not lift our pen. Thus, the function f is continuous on (0+∞).

Example. Using the graph, determine if the function h is continuous on the following intervals:
a. (-1, 1) b. [0.5, 2]

Solution:
a. If we trace the graph of the function h from the right side of x = −1 to the left side of x = 1, we
will be interrupted by a hole when we reach x = 0. We are forced to lift our pen just before we
reach x = 0 to indicate that h is not defined at x = 0 and continue tracing again starting from the
right of x = 0. Therefore, we are not able to trace the graph of h on (−1, 1) without lifting our
pen. Thus, the function h is not continuous on (−1, 1).

b. For the interval [0.5, 2], if we trace the graph from x = 0.5 to x = 2, we do not have to lift the
pen at all. Thus, the function h is continuous on [0.5, 2].
 Now, if a function is given without its corresponding graph, we must find other means to
determine if the function is continuous or not on an interval. Here are definitions that will help us:

A function f is said to be continuous...

a. everywhere if f is continuous at every real number. In this case, we also say f is continuous on
ℝ.
b. on (a, b) if f is continuous at every point x in (a, b).
c. on [a, b) if f is continuous on (a, b) and from the right at a.
d. on (a, b] if f is continuous on (a, b) and from the left at b.
e. on [a, b] if f is continuous on (a, b] and on [a, b).
f. on (a, 1) if f is continuous at all x > a.
g. on [a, 1) if f is continuous on (a, 1) and from the right at a.
h. on (−1, b) if f is continuous at all x < b.
i. on (−1, b] if f is continuous on (−1, b) and from the left at b.

Example. At what values of x is the function below continuous?

𝐶(𝑥) = 𝑥 5 − 2𝑥 4 + 8𝑥 3 − 5𝑥 2 + 7𝑥 − 10

Solution. Since the function C(x) is a polynomial function, then the function is continuous for all
values of x.

𝑥 2 +3𝑥+2
Example. When is the function R defined by 𝑅(𝑥) = continuous?
𝑥 2 −4

Solution. The rational function R becomes undefined when the denominator is zero which
happens when 𝑥 2 − 4 = 0 or 𝑥 = ±2. Thus, the rational function R is continuous for all values of x
except when x = 2 and x = -2.

Seatwork:
Determine if the polynomial or rational function is continuous at the given value of c. If the
function is continuous, fid the limit as x approaches c.

𝑥2
1. 𝑓(𝑥) = 2 − 5𝑥 3 ; 𝑐 = 1 6. 𝑓(𝑥) = ; 𝑐=1
𝑥 2 −𝑥
𝑥+4
2. 𝑓(𝑥) = 𝑥 2 − 3𝑥 − 1; 𝑐 = 1 7. 𝑓(𝑥) = ; 𝑐 = −4
𝑥 2 −16
𝑥
3. 𝑓(𝑥) = 𝑥 3 − 2𝑥 2 + 3𝑥 − 4; 𝑐 = 2 8. 𝑓(𝑥) = ; 𝑐=0
𝑥+1
3𝑥 𝑥 2 −3𝑥−4
4. 𝑓(𝑥) = ; 𝑐=1 9. 𝑓(𝑥) = ; 𝑐=0
𝑥+2 𝑥+1
𝑥 2 −9
5. 𝑓(𝑥) = ; 𝑐=2 10. 𝑓(𝑥) = 𝑥 3 + 3𝑥 − 1; 𝑐 = 1
𝑥−3
TYPES OF DISCONTINUITY

1. Removable Discontinuity
The function f(x) has a removable discontinuity at x = c if:
a. lim 𝑓(𝑥) exists; and
𝑥→𝑐
b. Either f(c)does not exist or f(c) ≠ lim 𝑓(𝑥).
𝑥→𝑐

Example. Let 𝑓(𝑥) = 𝑥 2 − 4. Show that the function has a removable discontinuity at x = 1.

Solution. We apply the three conditions of continuity to f(x) at x =1.

1. f(1) = 2
2. lim 𝑓(𝑥) = lim (𝑥 2 − 4) = 12 − 4 = −3
𝑥→1 𝑥→1
3. lim 𝑓(𝑥) ≠ 𝑓(1)
𝑥→1

Therefore, the function 𝑓(𝑥) = 𝑥 2 − 4 has a removable discontinuity at x = 1. Below is the

graph of the function with removable discontinuity.
 2. Essential Discontinuity
The function f(x) has an essential discontinuity at x = c if:
a. lim 𝑓(𝑥) does not exist.
𝑥→𝑐

Case I.
If lim− 𝑓(𝑥) exists and lim+ 𝑓(𝑥) exists but lim− 𝑓(𝑥) ≠ lim+ 𝑓(𝑥), then lim 𝑓(𝑥)
𝑥→𝑐 𝑥→𝑐 𝑥→𝑐 𝑥→𝑐 𝑥→𝑐
does not exist. This essential discontinuity is called jump discontinuity.

−2 𝑖𝑓 𝑥 < −2
Example. Show that 𝑓(𝑥) = {1 𝑖𝑓 − 2 ≤ 𝑥 ≤ 3 has a jump discontinuities at x = −2 and x =
5 𝑖𝑓 𝑥 > 3
3.

Solution. The graph of the function is illustrated below.

We have f(−2) = 1, with lim − 𝑓(𝑥) = −2 and lim + 𝑓(𝑥) = 1

𝑥→−2 𝑥→−2
Therefore, lim 𝑓(𝑥) does not exist.
𝑥→−2

We also have 𝑓(3) = 1, with lim− 𝑓(𝑥) = 1, and lim+ 𝑓(𝑥) = 5,

𝑥→3 𝑥→3
Therefore, lim 𝑓(𝑥) does not exist.
𝑥→3

Thus, f(x) has jump discontinuities at x = −2 and x = 3.

Case II.
If lim 𝑓(𝑥) = −∞ or lim 𝑓(𝑥) = +∞, then lim 𝑓(𝑥) does not exist. This essential
𝑥→𝑐 𝑥→𝑐 𝑥→𝑐
discontinuity is called an infinite discontinuity.

3𝑥−8
Example. Let 𝑓(𝑥) = . Show that f(x) has an infinite discontinuity at x = −4.
(𝑥+4)2
Solution. Note that 𝑓(−4) does not exist. Since the denominator is zero when x = −4 then we
obtain an infinite limit as x approaches −4.

3𝑥−8
We have lim − = −∞
𝑥→−4 (𝑥+4)2
Since lim − (3𝑥 − 8) = −20 and lim (𝑥 + 4)2 approaches 0 through positive
𝑥→−4 𝑥→−4−
values.

3𝑥−8
We also have lim + = −∞
𝑥→−4 (𝑥+4)2
Since lim + (3𝑥 − 8) = −20 and lim (𝑥 + 4)2 approaches 0 through positive
𝑥→−4 𝑥→−4+
values.

3𝑥−8
Therefore, lim 2 = −∞, which means that the limit of f(x) does not exist.
𝑥→−4 (𝑥+4)

Note that the function f(x) has an infinite discontinuity at x = −4 as shown in the graph
below.
FLOWCHART. Here is a flowchart which can help evaluate whether a function is continuous or not at
a point c. Before using this, make sure that the function is defined on an open interval containing c,
except possibly at c.
Name: ___________________________________________________________ Score:
____________________
Year & Section: ___________________________________________________ Date:
______________________

Exercises A:
Determine if the given function is continuous at a given value of x.

1. 𝑓(𝑥) = 3 − 𝑥; 𝑥 = 2 2. 𝑓(𝑥) = 𝑥 4 − 3𝑥 3 + 2𝑥 2 − 5𝑥 − 7; 𝑥 = −1

3𝑥
3. 𝑓(𝑥) = 𝑥 2 − 4𝑥 − 2; 𝑥 = 1 4. 𝑓(𝑥) = ; 𝑥=4
𝑥 2 −4

2𝑥
5. 𝑓(𝑥) = ; 𝑥=3 6. 𝑓(𝑥) = √3𝑥 − 10; 𝑥 = 2
𝑥−3

𝑥2 𝑥2
7. 𝑓(𝑥) = ; 𝑥=1 8. 𝑓(𝑥) = ; 𝑥 = −3
𝑥−2 𝑥−3

2
9. 𝑓(𝑥) = ; 𝑥=0 10. 𝑓(𝑥) = 6𝑥 − 6; 𝑥 = 0
𝑥
Exercise B.
Find the intervals on which each function is continuous.

𝑥 2 + 2𝑥 + 1, 𝑥 < 1 1, 𝑥 ≠ 1
1. 𝑓(𝑥) = { −𝑥 2. 𝑓(𝑥) = {
, 𝑥≥1 3, 𝑥 = 5
2

Find the intervals on which each function is continuous. You may use the provided graph to
sketch the function.
2𝑥 − 10, 𝑥 < 2 𝑥 2 −𝑥−2
3. 𝑓(𝑥) = { 4. 𝑓(𝑥) =
0, 𝑥 ≥ 2 𝑥+1

Exercise C.
1. Consider the graph of a function f(x) below.

a. Find all points where f(x) is not continuous.

b. What kind of discontinuity is each point?

2. Graph the different functions and identify the type of discontinuity.


 3 − x if x  2

a. f ( x) = 2 if x = 2
 x
 if x  2
 2

1 − x 2 if x  1
b. f ( x) = 
 2 if x  1

1 − x if x  3
c. f ( x) = 
3 − x if x  3