chapter 1 (Limits & Continuity)

Chapter 1 : Functions, Limits & Continuity

CHAPTER 1 : FUNCTIONS, LIMITS AND CONTINUITY

The concept of the Limits and Continuity is one of the most crucial things to understand in
order to prepare for calculus.

Who invented calculus?
Gottfried Leibnitz is a famous German philosopher and mathematician, and he was a
contemporary of Isaac Newton. These two gentlemen are the founding fathers of Calculus,
and they did most of their work in 1600s. We still use the Leibniz notation of dy for most

dx
purposes.

LIMITS

A limit is a number that a function approaches as the independent variable of the function
approaches a given value. For example, given the function f (x) = 3x , you could say, “The limit

of f (x) as x approaches 2 is 6 .” Symbolically, this is written lim f (x) = 6 .
x→2

Theorem of Limits

Properties Examples

a. lim c = c lim 2 = 2

x →a x →a

b. lim x n = an 1. lim x = 2

x→a x →2

2. lim x 3 = 23 = 8

x→2

3. lim x = +

x→+

4. lim x = −

x→−

lim [x 2 + x] = lim x 2 + lim x
x →3 x →3 x →3
c. lim [f (x)  g(x)] = lim f (x)  lim g(x)
x →a x →a x →a = 32 + 3

= 12

lim x 2(3 + x) = lim x 2 • lim (3 + x)
x →1 x →1 x →1
d. lim [f (x) • g(x)] = lim f (x) • lim g(x)
x →a x →a x →a = 12(3 + 1)

=4

lim f (x) lim 2x = lim 2x

x→2

e. lim f (x) = x →a , lim g(x)  0 x →2 x 2 + 3 lim x 2 + 3

x →a g(x) lim g(x) x →a x→2

x →a =4
7

lim 2(x + 4) = 2 lim (x + 4)
x →3 x →3
f. lim [cf (x)] = c lim f (x)
x →a x →a = 2(7)

= 14

g. lim n f (x) = n lim f (x), lim f (x)  0 lim 4 2x 3 + 7x + 51 = 4 lim (2x 3 + 7x + 51)
x→2 x→2

x →a x →a x →a = 4 81

n = positive integer. =3

1

MAT183/ MAT421 : Calculus I

Methods of Evaluating Limits

i) Determinate Forms of Limits (Limits by Direct Substitution)

To find lim f (x) , we substitute x = a in the function. If the value x →a comes out to be
x →a

definite value, it is limit. That is lim f (x)= f (a) provided is exists.
x →a

ii) Indeterminate Forms of Limits (Cannot use Direct Substitution)

Be careful when a quotient is involved. If direct substitution of x = a while evaluating
f (x)
lim g (x ) gives 0 or  , then we cannot use direct substitution anymore.
0 
x →a

 lim f (x) = 0
g (x ) 0
x →a

In limits, let say after we use direct substitution, the final answer gives 0 ; then there
0

are 3 cases.

Case 1 : f(x) or g(x) are quadratic function – factorize, simplify, subs the value of x.

Case 2 : f(x) or g(x) are surd ( ) function – multiplying by their conjugate, expand and
simplify using (a + b)(a − b) = a2 − b2, simplify, subs the value of x.

Case 3 : f(x) or g(x) are trigonometric functions – use squeezing theorem*, subs the value
of x.

Note : when we substitute the value of x into variable x, no more limits expressions (to write)

*squeezing theorem states that :

i) lim sin x = 1
x→0 x

Or lim sin2x = 1, lim sin3x = 1, lim sin 4x = 1, etc…
x→0 2x x→0 3x x→0 4x

ii) lim 1− cos x = 0
x→0 x

note :

1. when we use squeezing theorem, if the denominator not equal to the angle of sine, we
need to adjust the denominator until it is exactly equal to the angle of sine

2. never adjust the angle of sine in order to be equal to the denominator.

2

Chapter 1 : Functions, Limits & Continuity

 lim f (x) =  (positive infinite limits)
g(x) 
x→+

Case 1 : f(x) or g(x) are polynomial – dividing each expression (numerator and denominator)
by the highest power of x from denominator, simplify, subs the value of x by 
and use a = 0 (where a = constant) which means : any numbers divide by

 will approach to zero.

Case 2 : f(x) or g(x) are surd ( ) function – dividing each expression (numerator and

denominator) by the highest power of x from denominator, write x = x 2 or x2

= x 4 (etc..), simplify using a = a , subs the value of x by  and use a
bb 

= 0 (where a = constant) which means : any numbers divide by  will approach

to zero.

 lim f (x) =  (negative infinite limits)
g (x ) 
x→−

Case 1 : f(x) or g(x) are polynomial – dividing each expression (numerator and denominator)
by the negative highest power of x from denominator, simplify, subs the value
of x by  and use a = 0 (where a = constant) which means : any numbers

divide by  will approach to zero.

Case 2 : f(x) or g(x) are surd ( ) function – dividing each expression (numerator and

denominator) by the negative highest power of x from denominator, write x =

x 2 or x2 = x 4 (etc..), simplify using a = a , subs the value of x by 
bb

and use a = 0 (where a = constant) which means : any numbers divide


by  will approach to zero.

3

MAT183/ MAT421 : Calculus I

Examples for lim  f (x)  = 0 ,
 ) 0
x→a  g(x

 Quadratic or polynomial Functions (factorize)

a) Evaluate lim 16 − 4x . How about lim 16 − 4x ?
x→4 x2 − x − 12 x→0 x2 − x − 12

Solution :

Using direct substitution, Using direct substitution,

16 − 4 0 16 − 4 16 4 ( )
= = −3
2 − − 12 = 0 ( ) 2 − − 12 −12
→0
→4

( )

Thus, or,

16 − 4 4(4 − ) 16 − 4 = 4(4 − )

lim 2 − − 12 = ( + 3)( − 4) →0 2 − − 12 →0 ( + 3)( − 4)

→4 →4

−4( − 4) = −4( − 4)

= ( + 3)( − 4) →0 ( + 3)( − 4)

→4

−4 = −4

= ( + 3) →0 ( + 3)

→4

−4 −4
= (4 + 3) = (0 + 3)

4 4
= −7 # = −3 #

4

b) Evaluate lim x 2 + 3x − 10 Chapter 1 : Functions, Limits & Continuity
x→−5 x + 5
(ans : −7)
Solution :
(ans : 1 )
Using direct substitution, 2

2 + 3 − 10 0 5
= 0 ( )
→−5 + 5

( )

Thus,

2 + 3 − 10 = ( − 2)( + 5)

→−5 + 5 →−5 + 5

= ( − 2)

→−5

= (−5 − 2)

= −7 #

c) Evaluate lim x2 −4
x 2 + 4x − 12
x→2

Solution :

Using direct substitution,

2 − 4 = 0 ( )

→2 2 + 4 − 12 0

( )

Thus,

2 − 4 = ( + 2)( − 2)

→2 2 + 4 − 12 →2 ( − 2)( + 6)

= ( + 2)

→2 ( + 6)

(2 + 2)
= (2 + 6)

4
=8

1
=2 #

MAT183/ MAT421 : Calculus I

d) Evaluate lim 2t 3 + 3t 2 (ans : − 3 )
3t 4 − 2t 2 2
t →0

Solution :

Using direct substitution,

2 3 + 3 2 0
= ( )
3 4 − 2 2 0
→0

( )

Thus,

2 3 + 3 2 = 2(2 + 3)

→0 3 4 − 2 2 →0 2(3 2 − 2)

= (2 + 3)

→0 (3 2 − 2)

(0 + 3)
= (0 − 2)

3
= −2 #

e) Evaluate lim e2x − ex (ans : −1)
1− ex
x→0

Solution :
Note : After direct substitution, the answer gives 00. There is no surd and no trigonometric
functions → need to factorize.

2 − ( )2 − ( ) = =
1 − 1− Where =
lim  = lim  Factorize

→0 →0 −( − 1) = − + 1 = ( − 1)
Replace by
= lim  2 −
1 −
→0

= lim  ( − 1)
1 −
→0

= lim  ( − 1)
−( − 1)
→0

= −lim 

→0

= −lim 

→0

= − 0

= − 1  #

6

Chapter 1 : Functions, Limits & Continuity

 Surd (multiplying with their conjugate) Use (a + b)(a − b) = a2 − b2

What is Conjugate?? And why we multiplying with their conjugate??

expression conjugate multiplying with their conjugate

x −2 conjugate x +2 Multiplying with ( )( ) ( )x − 2 x + 2 = x 2 − (2)2 = x − 4
x −3 their conjugate Free from surd symbol
x +3 conjugate
Multiplying with ( )( ) ( )x + 3 x − 3 = x 2 −(3)2 = x − 9
their conjugate Free from surd symbol

conjugate Multiplying with ( )( ) ( )x + 3 − 2 x + 3 + 2 = x + 3 2 − (2)2

x+3 −2 x + 3 + 2 their conjugate = x+3−4
= x −1

Free from surd symbol

Examples
a) Evaluate lim x − 4

x→4 x − 2
Solution :

− 4 = − 4 ∙ √ + 2
→4 √ − 2 →4 √ − 2 √ + 2

= ( − 4)(√ + 2)
→4 (√ − 2)(√ + 2)

= ( − 4)(√ + 2)
(√ )2 − (2)2
→4

= ( − 4)(√ + 2)
− 4
→4

= (√ + 2)

→4

= √4 + 2

=2+2

=4 #

7

MAT183/ MAT421 : Calculus I

b) Evaluate lim x + 3 − 2
x→1 1− x

Solution :

√ + 3 − 2 = √ + 3 − 2 • √ + 3 + 2
1 − 1 − √ + 3 + 2
→1 →1

(√ + 3 − 2)(√ + 3 + 2)
=

→1 (1 − )(√ + 3 + 2)

= (√ + 3)2 − (2)2
→1 (1 − )(√ + 3 + 2)

+ 3 − 4
=

→1 (1 − )(√ + 3 + 2)

− 1
=

→1 (1 − )(√ + 3 + 2)

−(1 − )
=

→1 (1 − )(√ + 3 + 2)

−1
=

→1 √ + 3 + 2
−1

=
√1 + 3 + 2
−1

=
√4 + 2
−1

=2+2
1

= −4 #

8

Chapter 1 : Functions, Limits & Continuity

c) Evaluate lim x 2 + 8 − 3
x→−1 x + 1

Solution :

√ 2 + 8 − 3 = √ 2 + 8 − 3 √ 2 + 8 + 3
•
→−1 + 1 →−1 + 1 √ 2 + 8 + 3

(√ 2 + 8 − 3)(√ 2 + 8 + 3)
=

→−1 ( + 1)(√ 2 + 8 + 3)

(√ 2 + 8)2 − (3)2
=

→−1 ( + 1)(√ 2 + 8 + 3)

2 + 8 − 9
=

→−1 ( + 1)(√ 2 + 8 + 3)

2 − 1
=

→−1 ( + 1)(√ 2 + 8 + 3)

( + 1)( − 1)
=

→−1 ( + 1)(√ 2 + 8 + 3)

− 1
=

→−1 √ 2 + 8 + 3
−1 − 1

=
√1 + 8 + 3
−2

=
√9 + 3
−2

=3+3
2

= −6
1

= −3 #

9

MAT183/ MAT421 : Calculus I

d) Evaluate lim 1 −1
x+4 2
x →0
x
Solution :

1 − 1 1 1 1
2 ( 2)
√ + 4 = • −
√ + 4
→0 →0

= 1 • 2 − √ + 4
()
→0
2√ + 4

= 2 − √ + 4
→0 2 √ + 4

= 2 − √ + 4 • 2 + √ + 4
→0 2 √ + 4 2 + √ + 4

= (2 − √ + 4)(2 + √ + 4)
→0 2 √ + 4(2 + √ + 4)

(2)2 − (√ + 4)2
=

→0 2 √ + 4(2 + √ + 4)

4 − ( + 4)
=

→0 2 √ + 4(2 + √ + 4)

4 − − 4
=

→0 2 √ + 4(2 + √ + 4)
−

=
→0 2 √ + 4(2 + √ + 4)

−1
=

→0 2√ + 4(2 + √ + 4)

−1
=

2√4(2 + √4)

−1
= 2 • 2(2 + 2)

−1
=2•2•4

1
= − 16 #

10

Chapter 1 : Functions, Limits & Continuity

 Trigonometric Function : use Squeezing theorem

lim sin x = 1 and lim 1− cos x = 0
x→0 x x→0 x

In general, (from lim sin x = 1) if the angle of sine is exactly equal to the denominator, we
x→0 x

still can use the squeezing theorem as follows,

lim sin x = 1
x→0 x
lim sin 2x = 1
x→0 2x
lim sin 3x = 1
x→0 3x
lim sin 4x = 1
x→0 4x
lim sin 5x = 1 etc..
x→0 5x

 How about if the angle of sine is not equal to the denominator such as lim sin 2x ?
x→0 x

The solution should be like this,

Common mistake

lim sin 2x = lim 2 sin 2x lim sin 2x = lim 2 sin x
x→0 x x→0 2x x→0 x x→0 x

= 2 lim sin 2x = 2 lim sin x
x→0 2x x→0 x

= 2(1) = 2(1)

=2# =2#

(this is correct and acceptable) (this is wrong and not acceptable)

Caution :
if the angle of sine is not equal to the denominator

• never adjust the angle of sine
• adjust denominator until the denominator is equal to the angle of sine

11

MAT183/ MAT421 : Calculus I

Example 1 :
Find lim sin 3x
x→0 x

Solution :

3 = 3 3

→0 →0 3

= 3 3

→0 3

= 3(1)

=3#

Example 2 :
Find lim sin 3x

x→0 sin 2x

Solution :

3 = 3 • 1⁄
2 2 1⁄
→0 →0

= ( 3 )
( 2 )
→0

= ( 3 )
( 2 )
→0



→0

= 3 ( 3 3 )
( 2 2 )
→0

2

→0

3(1)
= 2(1)

3
=2 #

12

Chapter 1 : Functions, Limits & Continuity

Example 3 :
Find lim tan 7x

x→0 sin 3x

Solution :

7 = ( 77 )
3 3
→0 →0

7 3 ]
= [( 7 ) ÷

→0

7 1
= [ • 3 ]
7
→0

7 1
= [ • 7 ]
3
→0

= 7 • 1

→0 3 →0 7

= 7 • ( 1 ) • 1
3 ( 1 ) 7
→0 →0

= ( 7 ) • 1
( 3 ) 7
→0 →0

= ( 7 ) • 1
( 3 ) 7(0)
→0



→0

= 7 ( 7 7 ) • 1
3 ( 3 3 ) (0)
→0



→0

7(1) 1
= 3(1) • 1

7
=3 #

13

MAT183/ MAT421 : Calculus I

Example 4 :
Find lim cot 5x

x→0 cot x

Solution :

5 = ( 55 )
( )
→0 →0

5
= ( ÷ )
5
→0

5
= ( • )
5
→0

5
= ( • 5 )

→0

= ( 5 ) •
( 5 )
→0
→0

= 5 • • ( 1 )
( ) ( 5 ) ( 1 )
→0 →0

= 5(0) • ( )
(0)
→0 ( 5 )

1 ( )
1
= • →0

( 5 )

→0

= 5 ( )

→0 ( 5 5 )



→0

1
= 5(1)

1
=5 #

14

Chapter 1 : Functions, Limits & Continuity

Example 5 :
Find lim sin 2x

x →0

Solution :
2 = 2(0)

→0

= (0)
=0#

Example 6 :
Find lim cos 3x

x →0

Solution :
3 = 3(0)

→0

= (0)
=1#

Example 7 :

Find lim x3 + 4 sin7x
x→0 x

Solution :

3 + 4 7 3 4 7
= ( + )

→0 →0

= ( 2 + 4 7
)
→0

= ( 2) + 4 7
( )
→0 →0

= 2 + 4 7
( )
→0 →0

= 02 + 4(7) 7
( 7 )
→0

= 0 + (28)(1)

= 28 #

15

MAT183/ MAT421 : Calculus I

Example 8 :
Find lim sin x

x→0 2 x

Solution :

= lim ∙ √
lim
→0 2√ →0 2√ √

= lim √
2(√ )2
→0

= lim √
2
→0

= lim (√2 ∙ Separate using
) theorem of limits
→0

= lim (√2 ) ∙ lim (
)
→0 →0

Use direct subs. (no Use squeezing
more limits) theorem

√0 (1)
= (2) ∙
= (0) ∙ (1)
=0#

16

Chapter 1 : Functions, Limits & Continuity

Example 9 :

Find 2− 2
3 2
→0

Solution :

2 − 2 2 2
= −
3 2 3 2 3 2
→0 →0 →0

1 1
= − ( ∙ )
3 3
→0 →0

1 1
= − ( ∙ )
3 3
→0 →0

11
= 3 − 3 (1 ∙ 1)

11
=3−3

=0#

Example 10 :

Find ( 2 )

→0 cos

Solution :

( 2 ) = ( (2 ) ∙ 1 )
cos
→0 →0

= (2 ) ∙ 1 Separate
the limits
→0 →0 cos
Use direct
= 1 (2 ) ∙ 1 substitution
2 ( 0) (no more
Use squeezing →0 1 limit)
theorem 2

= 1 ( 2 ) ∙ 1
2 (2 ) (1)
→0

= 1 (1) ∙ 1
2 (1)

1
= 2#

17

MAT183/ MAT421 : Calculus I

Limits to Infinity

Infinity is a very special idea. We know we can't reach it, but we
can still try to work out the value of functions that have infinity in
them.

One Divided by Infinity
Let's start with an interesting example.

Question: What is the value of 1 ?

∞

Answer: We Can Approach It!

So instead of trying to work it out for infinity (because we can't get a sensible answer), let's
try larger and larger values of x:

x 1

1 x
2 1.00000
4 0.50000
10 0.25000
100 0.10000
1,000 0.01000
10,000 0.00100
0.00010

Now we can see that as x gets larger, 1 tends towards 0


We are now faced with an interesting situation:

• We can't say what happens when x gets to infinity

• But we can see that 1 is going towards 0



•

We want to give the answer "0 “but can’t, so instead mathematicians say exactly what is
going on by using the special word” limits

The limit of 1 as x approaches Infinity is 0


18

Chapter 1 : Functions, Limits & Continuity

And write it like this:

In other words:

As x approaches infinity, then 1 approaches 0



When you see "limit", think "approaching"

It is a mathematical way of saying "we are not talking about when x=∞, but we know as x
gets bigger, the answer gets closer and closer to 0".

Theorem of Infinite Limits : lim 1   0,
 x n 
x→

where n = 1, 2, 3, …

Examples :

i) lim  1 = 1 0
 x  
x→

ii) lim  1 = 1 = 1 0
 x2  2 
x→

iii) lim  1 = 1 = 1 0
 x3  3 
x→

iv) lim 3 = 3 0
 x  
x→

v) lim 4 = 4 = 4 0
 x 2  2 
x→

vi) lim 2 = 2 = 2 0
 x5  5 
x→

19

MAT183/ MAT421 : Calculus I

Examples for lim  f (x)  = 
 g (x ) 
x → 

 Positive and Negative Infinite Limits involving polynomial functions – we need to

simplify the function by dividing the numerator and denominator by the highest power

of x that occurs in the denominator.

Dividing the numerator and Factorize the highest power of x
denominator by the highest power for numerator and denominator
of x from denominator

1.a) i. 1.a) i.

(4 5 − + 1) 4 5 − + 1 5(4 − 1 + 1 )
4 3 4 − 2 2 4(3 4 5
4 5 − + 1 =
3 4 − 2 2 = − 2 2)
(3 4 − 2 2) →∞ →∞
→∞ →∞ 4

4 5 1 = (4 − 1 + 1 5)
4 4 4 4
− + →∞
= (3 − 2 2)
3 4 4−2 42
→∞

4 − 1 + 1 = ∞(4 − 1 + 1 )
3 4 ∞4 ∞5

= 3 − 2 (3 − ∞22)
2
→∞

= 4(∞) − 1 + 1 ∞(4 − 0 + 0)
∞3 ∞4 = (3 − 0)

3 − 2 ∞(4)
∞2 =3

4(∞) − 0 + 0 =∞ #
= 3−0

4(∞)
=3

= −∞ #

1.a) i. 4 5 − + 1 4 5 Take the term containing the
highest power of x for numerator
3 4 − 2 2 = 3 4
and denominator
→+∞ →+∞

4 5
=
3 4
→+∞

= 4

3 →+∞

= 4 (∞)
3

=∞ #

20

Chapter 1 : Functions, Limits & Continuity

Dividing the numerator and Factorize the highest power of x
for numerator and denominator
denominator by the highest power
of x from denominator

1.a) ii. 1.a) ii.

(4 5 − + 1) 4 5 − + 1 5(4 − 1 + 1 5)
− 4 3 4 − 2 2 4(3 4 2 2)
4 5 − + 1 = =
3 4 − 2 2 −
→−∞ →−∞ (3 4−− 42 2 →−∞ →−∞

)

4 5 1 = (4 − 1 + 1 5)
− 4 − 4 − 4 4
− + →−∞
= (3 − 2 2)
−3 44−−2 24
→−∞

−4 + 1 − 1 −∞ (4 − 1 + (−∞1 )5)
3 4 (−∞)4
=
= (3 − (−∞2 )2)
−3 + 2
→−∞ 2

−4(−∞) + 1 − 1 = −∞(4 − 1 − ∞15)
(−∞)3 (−∞)4 ∞4
=
−3 + 2 (3 − ∞22)
(−∞)2

−4(−∞) − 0 − 0 −∞(4 − 0 − 0)
= −3 + 0 = (3 − 0)

−4(−∞) −∞(4)
= −3 =3

= −∞ # = −∞ #

1.a) ii. 4 5 − + 1 4 5 Take the term containing the
highest power of x for numerator
3 4 − 2 2 = 3 4 and denominator

→−∞ →−∞

= 4 5

3 4
→−∞

= 4

3 →−∞

= 4 (−∞)
3

= −∞ #

21

MAT183/ MAT421 : Calculus I

Dividing the numerator and Take the term containing the
denominator by the highest power highest power of x for numerator
of x from denominator and denominator

1.b) 1.b)

4 2 + 1 3 4 2 + 1 3 4 2 3
(3 1) (2 ) = (3 1) (2 2 ) = (3 ) (2 2)
+ 2 + →−∞ + +
→−∞ →−∞

= 1 (2)3
(3)
→−∞

8
= (3)

→−∞

8
=3 #

• Never use Quick Method to solve limits
x approaches to any real number

• Quick method only can use to solve
Limits at Infinity

22

Chapter 1 : Functions, Limits & Continuity

1.c) page 41/ OCT 2007/MAT183/Q1a.i)/4 marks

Let h(x) = 17 + 2x , find lim h(x) + lim h(x) .
x 2 − 2x + 1 x→2 x→−

Solution :

ℎ( ) = 17 + 2

→2 →2 2 − 2 + 1

17 + 2(2)
= (2)2 − 2(2) + 1

17 + 4
=4−4+1

= 21

ℎ( ) = 17 + 2

→−∞ →−∞ 2 − 2 + 1

= (1 7 + 2)

→−∞ 2(1 − 2 + 1 2)


= 17 + 2

→−∞
(1 − 2 + 1 2)


= 17 + 2
−∞

−∞ (1 − 2 + (−∞1 )2)
−∞

−0 + 2
= −∞(1 + 0 + 0)

2
= −∞

=0

Hence,

ℎ( ) + ℎ( ) = 21 + 0
→2 →−∞

= 21 #

23

MAT183/ MAT421 : Calculus I

1.d) page 7/ OCT 2004/MAT183/Q1a /5 marks

Let h(x) = x3 − 9x lim h(x) − lim h(x) .
x3 − 3x 2 , find x→3 x→−

Solution :

ℎ( ) = 3 − 9

→3 →3 3 − 3 2

= ( 2 − 9)

→3 2( − 3)

= ( 2 − 9)

→3 ( − 3)

= ( + 3)( − 3)

→3 ( − 3)

= ( + 3)

→3

(3 + 3)
=3

6
=3

=2

ℎ( ) = 3 − 9

→−∞ →−∞ 3 − 3 2

= 3(1 − 92)
3(1 − 3 )
→−∞

= 1 − 9
2
→−∞
1 − 3


1 − 9
(−∞)2
=
1 − 3
−∞

1−0
=1+0

=1

Hence,

ℎ( ) − ℎ( ) = 2 − 1
→3 →−∞

=1 #

24

Chapter 1 : Functions, Limits & Continuity

1.e) page 35/ APR 2007/MAT183/Q1a(i) /5 marks

Let g(x) = x2 + x − 2 lim g(x) + lim g(x) .
x 2 − x , find x→1 x→+

Solution :

( ) = 2 + − 2

→1 →1 2 −

= ( − 1)( + 2)

→1 ( − 1)

= ( + 2)

→1

(1 + 2)
=1

=3

( ) = 2 + − 2

→+∞ →+∞ 2 −

= 2(1 + 1 − 2 2)
2(1 − 1 )
→+∞

= (1 + 1 − 2 2)
(1 − 1 )
→+∞

= 1 + 1 − 2
∞ ∞2

1 − 1
∞

1+0−0
= 1−0

=1

Hence,

( ) + ( ) = 3 + 1
→1 →+∞

=4 #

25

MAT183/ MAT421 : Calculus I

 Positive and Negative Infinite Limits involving Radicals
2.a) lim 25x 6 + 5

x→ x 6 − 11x3

Solution :

→ ∞ √ 2 65− 61+1 5 3 = √ 25 6 + 5 → ∞ √ 2 65− 61+1 5 3 = √ 25 6 + 5
6 − 11 3 6 − 11 3
→∞ →∞

(25 66+ 5) √ 6 (25 + 5 )
( 6 − 16 1 3) 6
= √ = →∞
6 (1 − 1 13)
→∞

25 6⁄ 6 + 5⁄ 6 = √ (25 + 5 6)
6⁄ 6 − 11 3⁄ 6 (1 − 1 13)
= √ →∞

→∞

25 + 5⁄ 6 = √25
1 − 11⁄ 3 1
= √

→∞ = √25
=5#
= √215−+151⁄⁄∞∞63

= √125−+151⁄⁄∞∞

= √25 + 0 → ∞ √ 2 65− 61+1 5 3 = √ 25 6
1−0 6
→∞

= √25 = √ → ∞ 25
=5#
= √25
=5#

26

Chapter 1 : Functions, Limits & Continuity

2. ) ( ) √2 2 − 3 2. ) ( ) √2 2 − 3

→+∞ + 2 →−∞ + 2

Solution : Solution :

√2 2 − 3 = √ 2(2 − 3 2) √ 2(2 − 3 2)

→+∞ + 2 →+∞ (1 + 2 ) √2 2 − 3 =

→−∞ + 2 →−∞ (1 + 2 )

√ 2 √2 − 3
2
= √ 2√2 − 3
(1 + 2 ) 2
→+∞ =

→−∞ (1 + 2 )

| |√2 − 3 | |√2 3
2 2
= −
(1 + 2 )
→+∞ =

→−∞ (1 + 2 )

√2 − 3 3
2 2
= − √2 −
2 )
→+∞ (1 + =

→−∞ (1 + 2 )

√2 − 3 3
2 2
= −√2 −
2 )
→+∞ (1 + =

→−∞ (1 + 2 )

√2 − 3 3
(∞)2 (−∞)2
= (1 + ∞2 ) −√2 −

√2 − 0 = (1 + −2∞)
(1 + 0)
= −√2 − 3
∞2
= (1 − ∞2 )
√2
=1 = −√2 − 0
= √2 # (1 − 0)

−√2
=1
= −√2 #

≥
ඥ = | | =

− <

27

MAT183/ MAT421 : Calculus I

2. ) ) √3 4 + 2 2. ) ) √3 4 + 2

→+∞ 4 2 − 3 →−∞ 4 2 − 3

Solution : Solution :

√3 4 + 2 √ 4(3 + 2 4) √3 4 + 2 = √ 4(3 + 2 4)

4 2 − 3 = 2(4 − 3 2) →−∞ 4 2 − 3 →−∞ 2(4 − 3 2)

→+∞ →+∞

√ 4√3 + 2 √ 4√3 + 2
4 4
=
= 2(4 − 3 2)
2(4 − 3 2) →−∞
→+∞

| 2|√3 + 2 | 2|√3 + 2
4 4
=
= 2(4 − 3 2) 2(4 − 3 2)
→−∞
→+∞

2√3 + 2 2√3 + 2
4 4
=
= 3 2) 2(4 − 3 2)
2(4 − →−∞
→+∞

√3 + 2 √3 + 2
4 4
=
= 3 2) (4 − 3 2)
(4 − →−∞
→+∞

√3 + 2 √3 + 2
(∞)4 (−∞)4
=
= (4 − (−∞3 )2)
(4 − (∞3)2)

= √3 + 0 √3 + 2
(4 − 0) ∞4
= (4 − ∞32)
√3
=4 # √3 + 0
= (4 − 0)

= √3 #
4

√ = ห ห = − ∞ < < +∞

28

Chapter 1 : Functions, Limits & Continuity

2.d) page 35/ APR 2007/ MAT183/ Q1a(ii)/ 5 marks
Find √3 2− .

→−∞ 4 −5

Solution :

√3 2 − = √ 2(3 − 1 )

→−∞ 4 − 5 →−∞ (4 − 5 )

√ 2√3 − 1 ≥
ඥ = | | =
=
(4 − 5 ) − <
→−∞

| |√3 − 1

=
(4 − 5 )
→−∞

− √3 − 1

=
(4 − 5 )
→−∞

−√3 − 1

=
(4 − 5 )
→−∞

−√3 − 1
−∞
= (4 − −5∞)

−√3 + 1
∞
= (4 + ∞5 )

= −√3 + 0
(4 + 0)

√3
=− 4 #

29

MAT183/ MAT421 : Calculus I

2.e) page 119/ MAR 2012/ MAT183/ Q1a(iii)/ 4 marks
Find lim 4x 8 + 3 . Ans : 2

x→ x8 − 6x

Solution :

→ ∞ √4 8 8−+6 3 8(4 + 3 8)
= √ − 6 7)
→∞ 8(1

= 4 + 3
√ − 8
→∞ 1
6
7

= 4 + 3
√ − ∞8

1 6
∞7

= √4 + 0
1 − 0

= √4
=2#

Two Sided limits

 If lim f (x) = L and lim f (x) = L then lim f (x) = L (exist)
x→a− x→a+ x→a

Lim x to a from the right

Or lim f (x) = L exist, if and only if lim f (x) = lim f (x)
x→a x→a− x→a+

Lim x to a from the left

 lim f (x)  lim f (x), then lim f (x) does not exist
x→a− x→a+ x→a

30

Chapter 1 : Functions, Limits & Continuity

Example 1 : f(x) single function (limit always exist)

Given f(x) = x 2 − 9 . Determine whether ( ) exist or does not exist.
x −3 →3

Solution :

( ) = 2 − 9 = ( + 3)( − 3) = ( + 3) = (3 + 3) = 6

→3− →3− − 3 →3− − 3 →3−

( ) = 2 − 9 = ( + 3)( − 3) = ( + 3) = (3 + 3) = 6

→3+ →3+ − 3 →3+ − 3 →3+

( ) = ( ) = 6
→3− →3+

ℎ , ( ) = 6 ( )

→3

Example 2 : g(x) piece-wise function  (limit always exist)

2−9 , ≠ 3
Given ( ) = { −3
.

4 , = 3

Determine whether ( ) exist or does not exist.

→3

Solution :

( ) = 2 − 9 = ( + 3)( − 3) = ( + 3) = (3 + 3) = 6

→3− →3− − 3 →3− − 3 →3−

( ) = 2 − 9 = ( + 3)( − 3) = ( + 3) = (3 + 3) = 6

→3+ →3+ − 3 →3+ − 3 →3+

( ) = ( ) = 6
→3− →3+

ℎ , ( ) = 6 ( )

→3

31

MAT183/ MAT421 : Calculus I

Example 3 : h(x) piece-wise function < > (limit does not exist)

2−9 , ≤ 3
Given ℎ( ) = { −3
.

4, > 3

Determine whether ℎ( ) exist or does not exist.

→3

Solution :

ℎ( ) = 2 − 9 = ( + 3)( − 3) = ( + 3) = (3 + 3) = 6

→3− →3− − 3 →3− − 3 →3−

ℎ( ) = 4 = 4
→3+ →3+

ℎ( ) ≠ ℎ( )
→3− →3+

ℎ , ℎ( )

→3

Example 4 : p(x) piece-wise function < > (limit exist)

2−9 , ≤ 3
Given ( ) = { −3
.

2 , > 3

Determine whether ( ) exist or does not exist.

→3

Solution :

( ) = 2 − 9 = ( + 3)( − 3) = ( + 3) = (3 + 3) = 6

→3− →3− − 3 →3− − 3 →3−

( ) = 2 = 2(3) = 6
→3+ →3+

( ) = ( ) = 6
→3− →3+

ℎ , ( ) = 6 ( )

→3

32

Chapter 1 : Functions, Limits & Continuity

CONTINUITY

Continuity is another far-reaching concept in calculus. A function can either be continuous or
discontinuous. One easy way to test for the continuity of a function is to see whether the graph
of a function can be traced with a pen without lifting the pen from the paper. For the math that
we are doing in precalculus and calculus, a conceptual definition of continuity like this one is
probably sufficient, but for higher math, a more technical definition is needed. Using limits, we’ll
learn a better and far more precise way of defining continuity as well. With an understanding
of the concepts of limits and continuity, you are ready for calculus.

What is Continuity ?

In Calculus, a function is continuous at x = a if and only if all three of the following conditions
are met :

1. ( ) is defined

2. ( ) exist or ( ) = ( )
→ → − → +

3. ( ) = ( )

→

If any one of the conditions is false, then we say that f is discontinuous at x = a, or that it
has a discontinuity at x = a.

There are two types of question involving continuity,
 test the continuity (use all the three conditions, step by step)
 determine the value of constant(s). (usually, we use

( ) = ( )
→ − → +

33

MAT183/ MAT421 : Calculus I

Example 1 :

Given f (x) = x 3 − 2x +1 , x2
3x −2 , x 2

Test whether f(x) is continuous at x = 2.

Solution :

Condition 1 : f (2) must defined

From f (x) = x3 − 2x + 1
 f (2) = (2)3 − 2(2) + 1 = 5 (defined)

Condition 2 : lim f (x) must exist
x→2

( )lim f (x) = lim x3 − 2x + 1 = (2)3 − 2(2) + 1 = 5
x →2− x →2−

lim f (x) = lim (3x − 2) = 3(2) − 2 = 4
x →2+ x →2+

Since lim f (x) ≠ lim f (x),  lim f (x) does not exist.
x →2− x →2+ x→2

So, no need to proceed with the third condition. And since one of the conditions is fail, thus the

conclusion is,

 The function is discontinuous at x = 2.

Extra notes :
Additional explanation using graphic representation :

the graph is disconnected
at x = 2

34

Chapter 1 : Functions, Limits & Continuity

Example 2 :

Given g (x ) = x 2 + x , x  −1
x + 1 , x  −1

Test whether g(x) is continuous at x = −1.

Solution :

Condition 1 : g(− 1) must defined
From g(x) = x + 1
 g(− 1) = − 1+ 1 = 0 (defined)

Condition 2 : lim g(x) must exist
x →−1

( )lim g(x) = lim x 2 + x = (− 1)2 + (− 1) = 0
x →−1− x →−1−

lim g(x) = lim (x + 1) = − 1+ 1 = 0
x →−1+ x →−1+

Since lim g(x) = lim g(x) = 0,
x →−1− x →−1+

 lim g(x) = 0 (exist)
x →−1

So, we need to proceed with the third condition.

Condition 3 : lim g(x) = g(− 1) must exist
x→−1

 lim g(x) = g(− 1) = 0
x →−1

Since all the three conditions are met, thus the conclusion is,
 The function is continuous at x = −1.

35

MAT183/ MAT421 : Calculus I

Example 3 :

x2 , x  3

Given f (x) = 2x +3 , 3  x  4

11 , x  4

Test whether f(x) is continuous at everywhere.

Solution : Notes:
• Continuous at everywhere means we test
 test the continuity at x = 3 continuity on x = 3 and x = 4 (boundary values)

Condition 1 : f (3) must defined • If the function continuous at all values of x, it says
From f (x) = 2x + 3 the function is continuous at everywhere.
 f (3) = 2(3) + 3 = 9 (defined)
• But, If the function is not continuous on one of the
x values, it says the function is discontinuous at
everywhere.

Condition 2 : lim f (x) must exist
x →3

lim f (x ) = lim x 2 = (3)2 = 9
x →3− x →3 −

lim f (x) = lim (2x + 3) = 2(3) + 3 = 9
x →3 + x→3+

Since lim f (x) = lim f (x) = 9,
x →3− x →3 +

 lim f (x) = 9 (exist)
x →3

So, we need to proceed with the third condition.

Condition 3 : lim f (x) = f (3) must exist
x →3
 lim f (x) = f (3) = 9
x→3

Since all the three conditions are met, thus the conclusion is,
 The function is continuous at x = 3.

36

Chapter 1 : Functions, Limits & Continuity

 test the continuity at x = 4

Condition 1 : f (4) must defined
From f (x) = 11
 f (4) = 11 (defined)

Condition 2 : lim f (x) must exist
x→4

lim f (x) = lim 2x + 3 = 2(4) + 3 = 11
x→4− x→4−

lim f (x) = lim 11 = 11
x→4+ x→4+

Since lim f (x) = lim f (x) = 11,
x→4− x→4+

 lim f (x) = 11 (exist)
x→4

So, we need to proceed with the third condition.

Condition 3 : lim f (x) = f (4) must exist
x→4
 lim f (x) = f (4) = 11
x→4

Since all the three conditions are met, thus the conclusion is,
 The function is continuous at x = 4.

Grand conclusion :
Since the function continuous for all values of x (including the boundary value x = 3 and
x = 4), thus the function is continuous at everywhere.

37

MAT183/ MAT421 : Calculus I

Example 4 :

5x + 7 , x  −1

Given h(x ) = 2x 2 + x +1 , −1 x  3

28 , x  3

Test whether h(x) is continuous at everywhere.

Solution :
 test the continuity at x = −1

Condition 1 : h(− 1) must defined

Use h(x) = 2x2 + x + 1
 h(− 1) = 2(− 1)2 + (− 1) + 1 = 2 (defined)

Condition 2 : lim h(x) must exist
x →−1

lim h(x) = lim (5x + 7) = 5(−1) + 7 = 2
x→−1− x →−1−

( )lim h(x) = lim 2x2 + x + 1 = 2(−1)2 + (−1)+ 1 = 2
x →−1+ x →−1+

Since lim h(x) = lim h(x) = 2,
x →−1− x →−1+

 lim h(x) = 2 (exist)
x →−1

So, we need to proceed with the third condition.

Condition 3 : lim h(x) = h(−1) must exist
x →−1
 lim h(x) = h(−1) = 2
x →−1

Since all the three conditions are met, thus the conclusion is,
 The function is continuous at x = −1.

38

Chapter 1 : Functions, Limits & Continuity

 test the continuity at x = 3

Condition 1 : h(3) must defined

Use h(x) = 2x2 + x + 1
 h(3) = 2(3)2 + (3) + 1 = 22 (defined)

Condition 2 : lim f (x ) must exist
x →3

( )lim h(x) = lim 2x2 + x + 1 = 2(3)2 + (3)+ 1 = 22
x →3 − x →3 −

lim f (x) = lim 28 = 28
x →3 + x →3 +

Since lim h(x) ≠ lim h(x),
x →3 − x →3 +

 lim h(x) does not exist
x→3

So, no need to proceed with the third condition. And since one of the conditions is fail, thus the
conclusion is,

 The function is discontinuous at x = 3.

Grand conclusion :
Since the function is not continuous on one of the x values, it says the function is discontinuous
at everywhere.

39

MAT183/ MAT421 : Calculus I

Example 5/ Page 170/ MAR 2016/ MAT183/ Q1b/ 9 marks

The function f(x) is defined as follows.

+ 1, ≤ 2
( ) = { 2 − 3, 2 < < 4

3 2 + 5, ≥ 4

2

i) Compute the value of k so that ( ) exists.

→2

ii) Hence, determine whether f is continuous at x = 4.

Solution :

i) Compute the value of k so that lim f (x) exists.
x→2

lim f (x) exists if and only if

x→2

lim f (x) = lim f (x) Note :
x →2− x →2+
• To find the constant k, we can
( ) lim (kx +1) = lim kx2 − 3 jump to condition 2.
x→2− x→2+
• No need to follow all the three
condition
(as mention page 34)

Using direct substitution (no more limits)

k(2) + 1 = k(2)2 − 3

2k + 1 = 4k − 3

3 + 1 = 4k − 2k

2k = 4
k=2#

ii) Hence, determine whether f is continuous at x = 4. Note :

• To test the continuity , we need
to follow all the three condition
(as mention page 34)

40

Chapter 1 : Functions, Limits & Continuity

Example 6/ Page 154/ SEP 2014/ MAT183/ Q1b/ 7 marks
The function g(x) is defined as follows.

+ 1 , < 0
( ) = { 2 2 , 0 ≤ ≤ 2

−2 + , > 2
i) Determine whether g(x) is continuous at x = 0.
ii) Find the value of k such that g(x) is continuous at x = 2.
Solution :

41

MAT183/ MAT421 : Calculus I

Example 7/ Page 53/ OCT 2008/ MAT183/ Q1c/ 9 marks

The function f(x) is defined as follows,

− 1, ≤ 1

( ) = {3 2 −+5 + 4, , 1 < ≤ 4
> 4
−4

Given that ( ) = 4,

→1

i) find the values of a and b.

ii) determine whether f(x) is continuous at x = 4.

Solution :

i) find the values of a and b.

Given that ( ) = 4,

→1

( ) = ( ) = 4
→1− →1+

where ( ) = 4 and ( ) = 4

→1− →1+

( − 1) = 4 (3 + ) = 4

→1− →1+

Using direct substitution (no more limits) Using direct substitution (no more limits)
(1) − 1 = 4 3(1) + = 4
= 5 # = 1 #

ii) determine whether f(x) is continuous at x = 4.

42

Chapter 1 : Functions, Limits & Continuity

Example 8/ APR 2008/ MAT183/ Q1b/ 5 marks

find the values of m and n if h(x) is continuous at x = 0.

, < 0

ℎ( ) = { = 0
,

+ 5, > 0

Solution :

We jump to the second condition,

h(x) is continuous at x = 0 if lim h(x) exist
x→0

thus, lim h(x) = lim h(x)
x →0 − x →0 +

where, lim h(x) = lim sinmx
x →0 − x→0− x

= m lim sinmx (using Squeezing theorem, adjust the denominator
x→0− mx so that it is equal to the angle of sine )

= m(1)

=m

And lim h(x) = lim (x + 5)
x →0+ x →0+

= 0 + 5 (using direct substitution, no more limits)

=5

from lim h(x) = lim h(x)
x →0 − x →0 +

 m=5#

To find n, we jump to the third condition,

h(x) is continuous at x = 0 if lim h(x) = h(0)
x →0

thus, lim h(x) = h(0) use h(x) = n
x →0 −  h(0) = n
lim sinmx = n
x→0− x
m =n
n =5#

END OF CHAPTER 1

43

MAT183/ MAT421 : Calculus I

Tutorial 1.1 : LIMITS

1. Evaluate the following limits. 5. Find

√5 2 + 2 Ans : 2 Ans :

) − 3 1. ) √5 ) 3 − 5 5. ) 0

→+∞ →∞

√2 + 1 − √3 ) 1 ) 2 ) 2
√3 3
) − 1 →0 3

→1 ) 1

2 − 2 ) 0 ) √4 − √6 − 16

) 3 2 →2 2 − 4

→0

2. Evaluate the following limits. 6. Evaluate each of the following limits.

√ 2 + 4 Ans : ) − 3 Ans :
2. ) 1
) 6 − 5 →3 3 − 2 − 6 1
6 15
→+∞ (2 ) 6. )
) 3
) √3 − 5 − 4 ) ) 1
8
→7 − 7 →0 2
) − 4
) 4 − 4 2 2 − ) 2
)
→0 3
→2 √ + 2 − √3 − 2

3. Evaluate each of the following limits. 7. Evaluate each of the following limits.

) 2 − 9 Ans : 4 − 2 Ans :
− 2 7. ) − 1
→3 2 2 − 7 + 3 6 )
5 ) − 1
5 − 3 3. ) →2
) 3
) 5 ) √ 2 + 8 − 3
→∞ √ 2 + 1 ) 2
→−1 + 1

) ( ) ) 0 2

→0+ 2√ )

→0

4. Evaluate the following limits. 8. Evaluate the following limits.

) 8 Ans : 2 − 1

→0 2 2 4. ) 4 ) −1 Ans :

→1

) √ + 5 − 3 ) 1 − 2 8. ) 2
2 − 8 12 ) ) 4
→4 ) 0
→2 2 − √6 −
2 − 7 + 10 ) ∞
) 2 − 1
)
→∞ − 5 2 + 3 + 5
→∞

44

Chapter 1 : Functions, Limits & Continuity

Tutorial 1. 2 : CONTINUITY

1. Sep 2011/MAT183/Q1b

Answer :
1. = 4, n= −2

g(x) is discontinuous at x = 3

2. Jun 2016/MAT421/Q1c Answer :
3. Apr 2010/MAT183/Q1b 2. = 152, b= −8

Answer :
3. i) k = −7

ii) f(x) is discontinuous at x = 2
iii) −∞

45

MAT183/ MAT421 : Calculus I Answer :
4. = −1, b= 1
4. Jun 2015/MAT421/Q1c
Answer :
5. Dec 2015/MAT421/Q1c 5. i) = −3

ii) f(x) is discontinuous at x = 1

6. Jun 2014/MAT421/Q1c Answer :
7. Jul 2017/MAT421/Q1c 6. = −1, b = −4
46
Answer :
7. k = 1

h(x) is continuous at x = 3