TYPES OF DISCONTINUITY

1. Removable Discontinuity – this discontinuity occurs when there is a hole in the graph of the
function.

Example:
𝑥 2 −4
1. Consider the function: f(x) = 𝑥−2

Solution:

The denominator of this rational function should not be equal to zero, that is x – 2 ≠ 0. If x – 2 ≠ 0, then
0
x = 2. Substituting 2 in place of x will make the rational function equal to 0 which is indeterminate. This
means that the function if discontinuous at x = 2. This can be verified by using the definition of
continuity.
𝑥 2 −4
A) f(x) = 𝑥−2
22 −4
f(2) = 2−2
𝟎
= 𝟎 or indeterminate
𝑥 2 −4 (𝑥+2)(𝑥−2)
B) lim 𝑥−2 = lim 𝑥−2
𝑥→2 𝑥→2
= lim x + 2
𝑥→2
= lim x + lim 2
𝑥→2 𝑥→2
=2+2
=4
C) lim f(x) ≠ f (2)
𝑥→2
Hence, the function is discontinuous at x = 2. There is a hole in the graph of the function.
f(x) = x + 2
X 0 1 2 3
f(x) 2 3 4 5 *insert graph*
ind.

The discontinuity of the function is removable. To remove the discontinuity, redefine the
function
𝑥2 − 4
𝑓(𝑥) = { 𝑥 − 2 , 𝑥≠2
4, 𝑥=2

𝑥 2 −4
F(x)= 𝑥−2
x 0 2
f(x) 2 4
*insert graph*
 2𝑥 2 +5𝑥−3
2. Consider the function: f(x) = 𝑥+3
2𝑥 2 +5𝑥−3
a) f(x) = 𝑥+3
2(−3)2 +5(−3)−3
f(-3) = (−3)+3
𝟎
= or indeterminate
𝟎
2𝑥 2 +5𝑥−3 (2𝑥−1)(𝑥+3)
b) lim = lim
𝑥→−3 𝑥+3 𝑥→−3 𝑥+3
= lim 2x - 1
𝑥→−3
= lim 2x - lim 1
𝑥→−3 𝑥→−3
= 2(-3) -1
= -7
c) lim f(x) ≠ f (-3)
𝑥→−3
Hence, the function is discontinuous at x = -3. There is a hole in the graph of the function.
f(x) = 2x - 1
X -3 -2 -1 0 1 2 3
f(x) -7 -5 -3 -1 1 3 5 *insert graph*
ind.

2𝑥 2 + 5𝑥 − 3
𝑓(𝑥) = { , 𝑥 ≠ −3
𝑥+3
−7, 𝑥 = −3

2𝑥 2 +5𝑥−3
f(x)= 𝑥+3
x 0 -3
f(x) -3 -7
*insert graph*
2. Jump Discontinuity – This discontinuity occurs when the graph of the function stops at one point and
seems to jump at another point. In a jump discontinuity, the left and the right hand limits exist but are
not equal.

Example:

1.
3𝑥 − 2, 𝑥 < −3
𝑓(𝑥) = {
𝑥 − 1, 𝑥 ≥ −3
Solution:

f(x) 3x -2

F -1 0 2 3
f(x) -5 -2 4 7

f(x) = x – 1

x 3 4 5 6
f(x) 2 3 4 5

Determine whether the functions are continuous at x = 3 using the definition of continuity.

A) f(x) = x – 1
f(3) = 3 – 1
=2
B) lim−3𝑥 − 2 = lim− 3𝑥 − lim−2
𝑥→3 𝑥→3 𝑥→3
= 3 (3) – 2
=7
C) lim−𝑥 − 1 = lim+ 𝑥 − lim+1
𝑥→3 𝑥→3 𝑥→3
=3–1
=2
D) lim−f(x) ≠ lim+f(x)
𝑥→3 𝑥→3

This is a non-removable discontinuity. It is not possible to redefine it. The f(x) is a jump
discontinuity.
 3. Infinite Discontinuity – In this type of discontinuity at least one of the limits infinite.

Example:
3
1. Consider the function f(x) = 𝑥−2

Solution:
3
A) F(x) = 𝑥−2
3
F(2) = (2)−2
3
= 0 or indeterminate
The first condition is not satisfied. Therefore, the function is discontinuous at x = 2.
B) Find the limits
x -4 -2 -1 0 1 1.5 1.7 1.9 1.99 1.999

f(x) -0.5 -0.75 -1 -1.5 -3 -6 -10 -30 -300 -3000

As the value of x approaches to 2 from the left, f(x) approaches negative infinity.

x 5 4 3 2.5 2.4 2.3 2.2 2.1 2.01 2.001

f(x) 1 1.5 3 6 7.5 10 15 30 300 3000

As the value of x approaches to 2 from the right, f(x) approaches positive infinity.