limitsExample: A function C=f(d) gives the number of classes C, a student takes in a day, d of the week. What does f(Monday)=4 mean? Solution. From f(Monday)=4, we see that the input day is Monday while the output value, number of courses is 4. Thus, the student takes 4 classes on Mondays.Function: is a rule which assigns an element in the domain to an element in the range in such a way that each element in the domain corresponds to exactly one element in the range. The notation f(x) read “f of x” or “f at x” means function of x while the notion y=f(x) means y is a function of x. The letter x represents the input value, or independent variable

1. www.covenantuniversity.edu.ng
Raising a new Generation of Leaders
OPEN DISTANT LEARNING CENTRE
COURSE MATRIAL
MAT121: Calculus III

2. 2
MODULE 2: LIMIT OF FUNCTIONS

3. 3
UNIT 1: LIMIT OF FUNCTIONS
Unit Objectives
After studying this unit, you should be able to:
❖define a limit;
❖evaluate the limit of a function by substitution;
❖evaluate the limit of a function by dividing out technique;
❖evaluate the limit of a function by rationalization;
❖compute one-sided limits;
❖determine if the limit of a function exists

4. 4
DEFINITION OF LIMIT OF FUNCTIONS
Let us consider the behaviour of the function
2
x -9
f(x)=
x - 3
around a point x=3.
X 2.9 2.99 2.999 2.9999 3.1 3.01 3.001 3.0001
f(x) 5.9 5.99 5.999 5.9999 6.1 6.01 6.001 6.0001
x 3
limf(x) 6.
→
=
You will notice in the Table, that as the values of x moves closer
to 3 from both directions, the values of the function f(x) moves
closer to 6. This means that the limit of the function f(x) as x
approaches 3 is 6. This is written as

5. 5
Definition:
If the values of f(x) can be made as close as we like to a unique
number L by taking values of x sufficiently close to a (but not equal
to a), then we write
which is read as “the limit of f(x) as x approaches a is L.”
lim ( )
x a
f x L
→
=

6. 6
Evaluating Limits Using Basic Limit Results
x a
lim x a
→
=
x a
limc c
→
=
x 4
lim x
→ x 3
lim10
→
x 4
lim x 4
→
=
x 3
lim10 10
→
=
1. Basic Limit Result: For any real number a and constant c,
i)
ii)
Example 2.1. Evaluate i)
.
ii)
Solution: Using i) you obtain
ii) Using (ii) you obtain

7. 7
Evaluating Limits Using Laws of Limits
7
x a
limf(x) L
→
=
x a
K
g
lim (x)
→
=
Let b and a be real numbers, let n be a positive integer, and
let f and g be functions with the following limits
and , then
x a
lim f(x) g(x)] L K
[
→
+ = +
x a
lim f(x) g(x)] L K
[
→
− = −
x a
lim f(x)g(x)] LK
[
→
=
ii. Sum Law:
iii. Difference Law:
iv. Product Law:
x a
lim f(x)] bL
[b
→
=
i. Scalar multiple Law:

8. 8
n n
x a
lim f(x)]
[ L
→
=
x a
f(x) L
lim
g(x) K
→
=
v. Quotient Law:
vi. Power Law:
n
n n
x a x a
lim f(x) limf(x) L
→ →
= =
L 0
 f(x) 0.

if n is even and
vii. Root Law:
for all L if n is odd and for

9. 9
9
x 5
lim(3x 3)
→−
+ ii)
2
3
x 2
2x 3x 1
lim
x 4
→
− +
+
Evaluate i)
x 5 x 5 x 5
lim(3x 2) lim 3x lim 2
→− →− →−
+ = +
x 5 x 5
2
x
3 lim lim
→− →−
=  +
3 ( 5) 2 17
=  − + = −
Solution. i) Applying ii
Applying i
ii)
( )
( )
2
2
x 2
3 3
x 2
x 2
lim 2x 3x 1
2x 3x 1
lim
x 4 lim x 4
→
→
→
− +
− +
=
+ +
Applying v
2
x 2 x 2 x 2
3
x 2 x 2
2 lim x 3 lim lim1
lim x lim4
x
→ → →
→ →
 −  +
=
+
( )
( )
2
x 2 x 2 x 2
3
x 2 x 2
2 lim x 3 lim x lim1
lim x lim4
→ → →
→ →
 −  +
=
+
2(4) 3(2) 1 1
(8) 4 4
− +
= =
+

10. 10
Computation of Limits by Substitution Method
Evaluate
Solution.
2
1
6
lim
3
x
x x
x
→−
+ −
+
2 2
1
6 ( 1) ( 1) 6
lim
3 ( 1) 3
1 1 6
1 3
6
3
2
x
x x
x
→−
+ − − + − −
=
+ − +
− −
=
− +
−
= = −

11. 11
Example: Evaluate
2
3
6 9
lim
3
x
x x
x
→
− +
−
Solution.
Factorize:
Substitute 3 for x:
3 3
2
2
3
6 9
l )
( 3
im lim
)
3
lim( 3
3
x x x
x
x
x
x
x
x
→ → →
− −
=
−
+
= −
−
3
lim( 3) 3 3 0
x
x
→
− = − =
Computation of Limits by Dividing through Technique

12. 12
Example: Evaluate
Solution.
0
1 1
lim
x
x
x
→
+ −
( )
( )
→ →
→
→
+
+
 
+ − + +
 =
 
+ + + +
 
+
=
+
−
+
=
0 0
0
0
1 1 1 1
lim lim
1 1 1 1
lim
1 1
lim
1 1
1 1
1
x x
x
x
x x
x x x x
x x
x
x
x
0
1 1 1 1
1 1 2
1
lim
1 1 0 1
x
x
→
= = =
+
+ +
+ +
Substituting 0 for x:
Computation of Limits by Rationalization Technique

13. 13
Evaluating Limits by Simplifying Complex Functions
Evaluate x 1
3 3
x 1 2
lim .
x 1
→
−
+
−
Solution. Simplifying yields
x 1 x 1 x 1
x 1 x 1 x 1
3 3
2(x 1
) 2(x 1
)
x 1 2
lim lim lim
x 1 x 1 x 1
2(x 1
)
li
6 3(x 1
) 6 3
)
m lim lim
x 1 2
3
(
x 3
3 3x
3x
1
3(x
(x ) x 1
) 2(x 1 x 1
)
)(
1
→ → →
→ → →
−
+ +
+ = =
− − −
+
= = =
− + −
− + − −
−
−
+ −
− −

14. 14
You now apply the dividing through technique to obtain
x 1 x 1
lim lim
2(x 1
)( )
3
2
(x 1
)
x 1
) x 1
3
(
→ →
=
+
−
− +
− −
Finally, you use the substitution method
x 1
3
lim
2(x 1
) 2(1 1
) 4
3 3
→
=
+
− −
= −
+

15. 15
Evaluating Limits When Laws of Limit Do Not Apply
Evaluate x 0
2 10
lim
x x(x 5)
→
 
+
 
−
 
Solution. First simplify the function
2
2 2
x 0 x 0 x 0 x 0
2 10 2x(x 5) 10x 2x 10x 10x 2
lim lim lim lim
x x(x 5) x 5
x (x 5) x (x 5)
→ → → →
− + − +
 
+ = = =
 
− −
− −
 
Finally, you apply the substitution technique
x 0 x 0
2 10 2 2
lim lim
x x(x 5) x 5 5
→ →
 
+ = = −
 
− −
 

16. 16
One-Sided Limits:
Right Hand Limit: If the values of f(x) can be made close as we like
to L by taking values of x sufficiently close to a (but greater than a),
then we write
Left Hand Limit: if the values of f(x) can be made close as we like to
L by taking values of x sufficiently close to a (but less than a), then we
write
lim ( ) .
x a
f x L
+
→
=
lim ( ) .
x a
f x L
−
→
=

17. 17
x 2
lim |x 2|
−
→
−
x 2
lim |x 2|
+
→
−
Evaluate i) ii)
Solution. i) When you approach x from the left of 2, you have
x 0
 . Subtracting 2 from both sides of the inequality gives
x 2 0
−  . This means |x 2|
− equals (x 2)
− − for x 2
 . You use
substitution to evaluate this limit
x 2 x 2
l 0
-
im |x 2| lim (x 2) (2 2)
− −
→ →
− = − = − − =
ii) Approaching x from the right of 2 gives x>0. Subtracting 2
from both sides gives x-2>0. This means |x 2|
− equals x-2 for
x<2. You use substitution to evaluate this limit
x 2 x 2
lim |x 2| lim(x 2) (2 2) 0
+ +
→ →
− = − = − =

18. 18
Example. Evaluate the following a) b)
0
|2 |
lim
x
x
x
−
→ 0
|2 |
lim
x
x
x
+
→
0 0
0
2
|2 |
2
lim l
) im
lim 2
x x
x
x
a
x x
x
− −
−
→ →
→
= −
= −
= −
0 0
0
|2 |
lim lim
li
2
m
2
)
2
x x
x
b
x
x
x
x
+ +
+
→ →
→
=
=
=
Solution.

19. 19
Existence of Limits
If f(x) is a function and a and L are real numbers, then
x a
limf(x) L.
→
=
if and only if both the left-hand and right-hand limits exist
and are equal to L.

20. 20
Discuss the existence of the limit of 2
4 if 1
( ) as 1
4 if 1
x x
f x x
x x x
− 

= →

− 

1 1
lim ( ) lim(4 )
4 1
3
x x
f x x
− −
→ →
= −
= −
=
1
2
2
1
3
l
1
im
4( ) (1)
( ) l
4
im4
1
x
f x x x
+ +
→ →
=
−
−
−
=
=
=
Since the left hand limit is equal to the right hand limit, the limit exists.

21. 21
Review
1. Evaluate following limits
2
5 3
1 1
i
3 2 1
lim
)
x
x x
x x
→−
−
+ −
−
2
3
6 9
ii) lim
3
x
x x
x
→
− +
− 0
1 1
iii) lim
x
x
x
→
+ −
2. Given the function , compute
2
| 2|( )
( )
2
x x x
f x
x
− +
=
−
2
m
i li
)
x +
→ 2
m
i li
i)
x −
→

22. 22
Answers
1i) 5/4 ii) 0 iii) 2
2i) -6 ii) 6