CHAPTER 1

LIMIT AND
1 CONTINUITY

OBJECTIVES
At the end of this chapter the students should be able to
• Define the concept of limit and limit notation
• Use basic rules of limit
• Solve one-side limit and infinite limit
• Determine the continuity using graph and point

INTRODUCTION
Consider the function below :
f ( x) = 4
x +1
The values of f(x) always change when the values of x change. If we want to know
the values of f(x) when we take the values of x approach 3 but are not equal to 3.

x 2 2.9 2.99 2.999 3 3.0001 3.001 3.01 3.1

F(x) 1.333 1.026 1.003 1.000 0.9999 0.999 0.998 0.975
2 8 7 6

From the table above, we learn that when x approaches 3, the values of f(x)
4 =1
approaches1. That means we can write lim
x→3 x +1

1.1 DEFINITION

Consider any function f(x) which is defined on any number approaching a but not
defined on a. When x approaches a, then f(x) will approach any number, called L and
we write
lim f ( x) = L
x→a

1.2 BASIC RULES OF LIMIT

1. lim L = L where L is constant

x→a
Example
lim 8 = 8
x →6

lim100 = 100
x →2

2. lim xn =a n
x→ a
Example
lim x 2 = 22 = 4
x→2
CHAPTER 1

3. lim  f (x ) ±g (x ) = lim f (x ) ±lim g (x )

x→ c
 
 x→ c x→ c
Example
( )
lim 2 x − x 2 = lim 2 x − lim x 2
x →1 x →1
x →1

= 2(1) − 12 = 1

 f ( x ) •g ( x )=lim f (x ) • lim g ( x )
4. xlim
→ c
 
 x→ c x→ c
Example

x →1
( )
lim 2 x • x 2 = lim 2 x • lim x 2
x →1 x →1

= 2(1) • 1 = 2
2

f (x ) =xlim
lim g → c
f (x )
5. x→ c (x ) lim g (x )
x→ c
Example

2 x lim 2x 2
lim = x →1
= =2
x →1 x 2 lim x 2 1
x →1

4. lim kf ( x ) =k lim f ( x ) where k is any real number

x→a  x→a
Example

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

x →1 x →1

1
 
5. lim n f (x ) =n lim f (x ) = lim  f ( x) n

x→ a x→ a x→ a




 
Example

lim
x→1
2 x −1 = lim(2 x −1) [ x→1
]
1
2

1
=[2(1) −1]2
1

=1 2
= 1 =1

Note:
CHAPTER 1
lim f ( x) = L ,
1. If
x→ a then the limit is exist, i.e L

2. If lim f ( x ) = 0 , then the limit is exist, i.e 0

x→a L

3. If lim f ( x) = L , then the limit is not exist

x→a 0

4. If lim f ( x) = 0 , to find the limit we should change f(x) to another form

x→a 0
with

i) factorising
ii) times with conjugate

Example:

1. lim 2 x = 2( 2 ) = 4 limit is exist = 4

x →2

2 x − 8 2( 4 ) − 8 0
2. lim = = =0 limit is exist = 0
x →4 3 3 3

1 1
3. lim = limit is not exist(dividing by 0 undefined
x →3 x − 3 0
4.a) factorising

x 2 − 2 x − 3 32 − 2( 3) − 3 0
lim = =
x→ 3 x−3 3− 3 0

We could have evaluated this limit by factorizing first:

b) Times with conjugate

x+4 −2 0+4 −2 0
lim = =
x→0 x 0 0

So
CHAPTER 1
x +4 −2 x +4 +2
lim •
x→0 x x +4 +2

Expand the numerator because we take the numerator as a conjugate

x+4 −2 x+4 +2
lim •
x→0 x x+4 +2

x+4+2 x+4 −2 x+4 −4

lim
x→0 (
x x+4+2 )
x
lim
x→0 x (x+4 +2 )
1
lim
x→0 x+4+2
1 1 1 1
= = =
0+4 +2 4 +2 2+2 4

1.3 ONE-SIDED LIMITS

Definition :

1. Let f any function which defined on open interval (a,c). Then the limit of
function f when x approaches a from the right is eqaul to L, we write
lim f ( x) =L
x→ a+
2. Let f any function which defined on open interval (b,a). Then the limit of
function f when x approaches a from the left is eqaul to L, we write
lim − f ( x) =L
x→ a
Theorem :
lim f ( x ) =L exist, if and only if
x→ a
lim f ( x ) = lim −f ( x ) =L
x→ a+ x→ a
(right-hand limit = left-hand limit)
E.g :
lim 5 x + 1 = 5( 2 ) + 1 = 11
x →2 +

lim 5 x + 1 = 5( 2 ) + 1 = 11
x →2 −
Hence limit is exist because right-hand limit = left-hand limit
lim 5 x + 1 = 5( 2 ) + 1 = 11
x→2
CHAPTER 1

e.g
x
Determine lim exist or not.
x →0 x

x x≥0
x =
− x x<0

x 1
lim+ = lim+ = 1
x→0 x x→0 1

−x
lim− = lim− − 1 = −1
x→0 x x→0

x
Hence lim not exist because right-hand limit ≠ = left-hand limit
x →0 x

1.4 INFINITE LIMIT

Definition
Let f(x) any function which defined on interval (a,+∞) . The limit of f(x) when x
increase without bound is equal to L, then we write

lim f ( x ) =L
x→ +∞
Definition
Let f(x) any function which defined on interval ( −∞, a ) . The limit of f(x) when x
decrease without bound is equal to L, then we write

lim f ( x) = L
x→ −∞
E.g :

lim x + 1 = −∞ + 1 = −∞
x → −∞

lim x 2 ( − x ) = ( + ∞ ) ( − ∞ ) = −∞
2
x → +∞

Theorem
If m is any positive number, then
CHAPTER 1
1
a) lim m =0
x →+∞x
1
b) lim =0
x→ − ∞xm

E.g :
1 1
lim = =0
x → +∞ x 2
+ ∞2

1 1
lim = =0
x → −∞ x 3
− ∞3

f ( x) ∞
Case 1: lim =
x → ±∞ g ( x) ∞
 1 
 n
f ( x)  x  where n is the highest
Step : Change the form of by multiplying it with
g ( x)  1 
 n
x 
power of x in g(x)

Example

1. Find the limit .

Answer:

 5 − 3∞  ∞
lim  =
 6∞ + 1  ∞
x → +∞

So using the formula we get

 5 3x 
 − 
 5 − 3x 
lim   = lim x x 
x → +∞ 6 x + 1
  x → ∞ 6 x + 1 
 
 x x

2. Find
CHAPTER 1
Answer

1.5 CONTINUITY
1. Determine the Continuity from Graph

The continuous function graph is the graph which no disconnect or has ‘hole’.

E.g :

a
Continuous function Not-
continuous function
because the graph
is
disconnect at x = a

x=a

Not-continuous function because the graph is not defined at x = a

2. Continuity at Point
CHAPTER 1
Definition
Any function f(x) is called continuous at point x = a if and only if it meet the
following three conditions below:

a) f(a) is defined
b) lim f ( x) exist i.e lim +f ( x) =x→
lim −f ( x)
a
x→ a x→ a
c) f (a) = lim f ( x)
x→ a
If any conditions above are not fulfilled then f(x) is not-continuous at x = a.

E.g :

Determine whether this function continuous or not at x = 4

2 , 0 < x ≤ 4
f ( x) = 
 x , 4< x≤8

Answer:

a) f ( a ) = f ( 4) = 2
lim f ( x) = lim x = 4 =2
b)
x→ a+ x →4 +
lim f ( x) = lim 2=2
x→ a+ x→ 4−

lim f ( x) =wujud =2
x→ 4

c) f 4  = lim f ( x) =2

x→4
hence f(x) continuous at x=4

EXERCISE 1.1

Determine limit for the following function using basic rules:

CHAPTER 1
lim lim 3 lim
a. 68. b. x − 7 x. c. x 2 + x.
x→2 x→2 x → −4

lim x +1 lim 3 lim x − 3

d. e. f.
x → −1 2 x − 3 x→4 4 x − 16 x → 3 x2 −1

EXERCISE 1.2

Determine limit for the following function using factorization and time with
conjugate:

lim x − 3 lim 9 − x 2 lim x−5

a. . b. . c. . 2
x → 3 x2 − 9 x →3 x−3 x → 5 x − 25

lim 4 − x lim −x −2 lim 3− x−4

d. . e. . f. .
x →4 2− x x → −4 x + 4 x → 7 7x − x2

EXERCISE 1.3

Solve the following infinite limit

lim 4 lim x 2 − 4 x − 1 lim 2 x 3 − 6 x + 1

a. 3+ 2 b. c.
x→∞ 2x x → ∞ 3x 2 + 4 x→∞ x+8

EXERCISE 1.4

1. Determine whether the following functions continuous or not at the

given point
 x + 1 jika x < 1 
 2 
a. f ( x) =  x − 3 x + 4 jika 1 ≤ x ≤ 3 x=1
5 − x jika x > 3 
 

8 if x = 4 
 
b. f ( x) =  16 − x 2  x=4
 if x ≠ 4
 4−x 

c. f ( x) = { x − 2 } x=2