Lec-7 (Applied Calculus) B.Sc.

Engineering 1st Semester

Continuity Of a Function at a Point

A function f(x) is continuous at x=c, c∈ 𝐷 f if and only if it meets the
following three conditions:
1. 𝑓(𝑐) exists.
2. lim 𝑓(𝑥 ) exist i.e lim− 𝑓(𝑥 ) = lim+ 𝑓(𝑥 ) = 𝑎
𝑥→𝑐 𝑥→𝑐 𝑥→ 𝑐
3. lim 𝑓(𝑥)=𝑓(𝑐)
𝑥→𝑐

Example 1
Discuss the continuity of
𝑥 −1 𝑥<3
𝑓(𝑥) = { at 𝑥 = 3
2𝑥 + 1 𝑥≥3
Solution:
𝑓(3) = 2(3) + 1 = 7
lim 𝑓 (𝑥 ) =?
𝑥→3

lim 𝑓(𝑥 ) = lim−(𝑥 − 1) = 3 − 1 = 2

𝑥→3− 𝑥→3

lim 𝑓 (𝑥 ) = lim+(2𝑥 + 1) = 7
𝑥→3+ 𝑥→3

∴ lim− 𝑓(𝑥) ≠ lim+ 𝑓(𝑥)

𝑥→3 𝑥→3

∴ lim 𝑓 (𝑥 ) Does not exist.

𝑥→3

∴ 𝑓 (𝑥 ) is discontinuous at 𝑥 = 3.
Example 2
Discuss the continuity of 𝑓(𝑥) = |𝑥 − 5|, 𝑎𝑡 𝑥 = 5.
Solution:
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Lec-7 (Applied Calculus) B.Sc. Engineering 1st Semester

𝑥 −5 𝑖𝑓 𝑥 − 5 ≥ 0
𝑓(𝑥) = |𝑥 − 5| = {
−(𝑥 − 5) 𝑖𝑓 𝑥 − 5 < 0
𝑥 − 5 𝑖𝑓 𝑥 ≥ 5
𝑓(𝑥) = |𝑥 − 5| = {
−(𝑥 − 5) 𝑖𝑓 𝑥 < 5

𝑓(5) = 5 − 5 = 0
lim 𝑓(𝑥 ) =?
𝑥→5

lim 𝑓(𝑥 ) = − lim−(𝑥 − 5) = −(5 − 5) = 0

𝑥→5− 𝑥→5

lim 𝑓(𝑥 ) = lim+(𝑥 − 5) = 5 − 5 = 0

𝑥→5+ 𝑥→5

∴ lim− 𝑓(𝑥 ) = lim+ 𝑓(𝑥)

𝑥→5 𝑥→5

∴ lim 𝑓(𝑥 ) = 0
𝑥→5

lim 𝑓 (𝑥 ) = 𝑓(5) = 0
𝑥→5

∴ 𝑓 (𝑥 ) is continuous at 𝑥 = 5
Example 3
Discuss the continuity of
𝑥2 −9
𝑥≠3
𝑓(𝑥 ) = { 𝑥−3 at 𝑥 = 3.
6 𝑥=3

Solution:
𝑓(3) = 6

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Lec-7 (Applied Calculus) B.Sc. Engineering 1st Semester

𝑥2 −9
lim 𝑓 (𝑥 ) = lim
𝑥→3 𝑥→3 𝑥−3

(𝑥+3) (𝑥−3)
=lim
𝑥→3 (𝑥−3)

=lim (𝑥 + 3) = 3 + 3 = 6
𝑥→3

lim 𝑓 (𝑥 ) = 𝑓(3) = 6
𝑥→3

𝑓(𝑥) is continuous at x=3

Definition:-
A function ‘f’ is continuous at left end- point x=a of
the interval[a,b],if lim+ 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎

Definition:-
A function ‘f’ is continuous at right end point x=b of
the interval [a,b],if
lim 𝑓(𝑥) = 𝑓(𝑏)
𝑥→𝑏 −

Definition:- A function ‘f’ is continuous in an interval [a,b] if it is

continuous for all 𝑥 [𝑎, 𝑏]

Theorem# 1:-
(continuity of algebraic combinations)
If functions ‘f’ and ‘g’ are continuous at x=c , the following
functions are also continuous at x=c
(1) f+g & f-g
(2) fg

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Lec-7 (Applied Calculus) B.Sc. Engineering 1st Semester

𝑓
(3) , provided g(c) ≠ 0
𝑔
(4) kf where ‘k’ is any constant
(5) [𝑓(𝑥 )]𝑚/𝑛 ,provided[𝑓(𝑥 )]𝑚/𝑛 is defined on an interval
containing “c” and m,n are integers.
Theorem # 2:-
Every polynomial is continuous at every point of real line.
Every rational function is continuous at every point where its
denominator is not zero.
𝑥+1
f(x) =
𝑥−1

continuous at every point except 1.

Types of discontinuity: There are two types of discontinuity
Non – Removable discontinuity:-
If at any point, a function does not have functional value as well
as limiting value then the type of discontinuity is called non
removable discontinuity.
1
e.g f(x)=
𝑥

f(0) does not exist.

lim 𝑓(𝑥 ) does not exist.
𝑥→0

Removable discontinuity:-
If at any point, a function does not have functional value but it has
limiting value at that point then that kind of discontinuity is called
removable discontinuity.

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Lec-7 (Applied Calculus) B.Sc. Engineering 1st Semester

𝑥2 −4
e.g f(x)= x 2
𝑥−2

f(2) does not exist

lim 𝑓(𝑥) = 4
𝑥→2

Continuous extension :-
If f(c) does not exist . but lim 𝑓(𝑥) = L then we can define a new
𝑥→𝑐
function 𝐹(𝑥) as
𝑓(𝑥 ) 𝑖𝑓 𝑥 ≠ 𝑐
F(x) = {
𝐿 𝑖𝑓 𝑥 = 𝑐

e.g.
𝑥2 −4
F(x) = { 𝑥−2 𝑥≠2
4 𝑥= 2
Then the function F(x) is continuous at x=c . It is called continuous
extension of f(x) to x=c.
Example # 4:-
𝑥2 +𝑥−6
Show that 𝑓 (𝑥 ) =
𝑥2 −4

has a continuous extension to x=2 . Also find that extension.

Solution:- 𝑓(2) does not exist.
𝑥2 +𝑥−6
lim 𝑓(𝑥) = lim
𝑥→2 𝑥→2 𝑥2 −4

(𝑥−2)(𝑥+3)
= lim (
𝑥→2 𝑥−2)(𝑥+2)

2+3
=
2+2
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Lec-7 (Applied Calculus) B.Sc. Engineering 1st Semester

5
=
4

Thus 𝑓(𝑥) has a continuous extension at x=2 which is given by

𝑥2 +𝑥−6
𝑥≠2
𝑥2 −4
𝐹(𝑥) = { 5
𝑥=2
4

Example # 5:-
Define g(3) in a way that extends 𝑔(𝑥) = (𝑥 2 − 9)/(𝑥 − 3) to be
continous at x =3.
Solution:
g(x) = (𝑥 2 − 9)/(𝑥 − 3)
g(3) does not exist .
lim g(x) = lim ( 𝑥 2 − 9)/(𝑥 − 3)
𝑥→3 𝑥→3

= lim(𝑥 − 3)(𝑥 + 3)/(𝑥 − 3)

𝑥→3

= lim(𝑥 + 3)=(3+3)=6
𝑥→3

Thus g(x) has a continues extension to x=3 which is given by

𝑥2 −9
g(x) = { 𝑥≠3
𝑥−3
6 𝑥=3
Example # 6:-

For what value of ‘a’

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Lec-7 (Applied Calculus) B.Sc. Engineering 1st Semester

𝑥2 − 1 𝑖𝑓 𝑥 < 3
f(x) = {
2𝑎𝑥 𝑖𝑓 𝑥 ≥ 3
is continuous at every ‘𝑥’.
Solution:-
Since 𝑓(𝑥) is continuous for every ‘𝑥’ .

It is continuous at x = 3

then
lim 𝑓 (𝑥 ) = lim+ 𝑓(𝑥)
𝑥→3− 𝑥→3

lim−(𝑥 2 − 1) = lim
+
(2𝑎𝑥)
𝑥→3 𝑥→3

9-1 = 2a(3)
8 = 6a
4
a =
3

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Lec-7 (Applied Calculus) B.Sc. Engineering 1st Semester

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