Removable Discontinuity 

Are you the student who gets stuck in graph questions? Regardless of how interesting and easy they may be, a lack of basic concept clarity will always lead you to dread them. However, acing the graph-based questions can be easy. Read on removable discontinuity, which is explained in simpler terms. Learn from the practice questions and then re-evaluate yourself. Let us begin!

What is Discontinuity? 

The continuous functions are important in mathematics owing to their essential applications. However, continuity is not always witnessed. A point or interval is often seen in the function, making it discontinuous. Hence, discontinuity can be defined as the graph’s jump, break or gap at a particular location. The open circle marks it and occurs when the function has a different value than the calculated one based on the equation. A discontinuity can be removable, asymptotic or jump. 

Removable discontinuity: The removable discontinuity is undefined at the point of concern. Since it is removable, it can be made continuous by redefining the function at that point. An open circle in the graph marks the removable discontinuity. 

Jump discontinuity: The jump discontinuity is marked by the gap as the function will jump from one point to another. 

Asymptotic Discontinuity: The asymptotic discontinuity is evident by the condition when the function approaches different values from either side of a particular point. It is represented by a graph approaching two horizontal asymptotes from opposite sides. 

How to Find Removable Discontinuity?

You must follow the stepwise procedure below to find the removable discontinuity.

Step 1: Factor both the numerator and denominator in the given function 

Step 2: Find the common factors in the results obtained in Step 1

Step 3: Equate the common factors with 0. It will give you the value of x. 

Step 4: Plot the graph and mark the point with a circle. (Note: You have got the results in the third step. Step 4 is for visualization of the removable discontinuity). 

Learn to Solve Removable Discontinuity with Examples

Q1. Find out whether the function f(x)=x2-5x+6/x-2 has removable discontinuity. 

We will follow the mentioned steps to identify the removable discontinuity in the given equation. It will be as follows: 

Factorization: f(x)=(x-2)(x-3)/x-2

Find common factors: f(x)=x-2

Set common factors to zero: x-2=0

x=2

Hence, the removable discontinuity of the function is at the point 2. 

Q2. Determine if the function f(x)=x2+2 is continuous at x=2. 

To find if the mentioned function is continuous at a stated point, we will have to evaluate the function at 2, and the limit of the function as x approaches 2. Here’s how it will be done: 

Function at x=2: 

f(2)=(2)2+2

f(2)=4+2

f(2)=6

Limit as x approaches 2: 

x2f(x)

x2(x2+2)

(22+2)

(4+2)

6

Since both function and limit are equal, the given function is considered continuous at x=2 

Q3. Find the removable discontinuity in the function: f(x)=x2-9/x2-5x+6.

We will follow the stepwise procedure to find the removable discontinuity in the given function: 

Factorization of numerator: (x+3)(x-3)

Factorization of the denominator: (x-2)(x-3)

Factorization: f(x)=(x+3)(x-3)/(x+2)(x-3)

Find common factors: The common factors obtained on the factorization is (x-3)

Sect common factors to zero: x-3=0

x=3

Hence, the removable discontinuity will occur at 3. 

Q4. Find the removable discontinuity in the function: f(x)=x2-16/x2-7x+10. 

Following the stepwise procedure to find the removable discontinuity in the given function: 

Factorization of numerator: (x+4)(x-4)

Factorization of the denominator: (x-2)(x-5)

Factorization: f(x)=(x+4)(x-4)/(x-2)(x-5)

Find common factors: There aren’t any common factors between the numerator and denominator. 

Hence, it can be said that the given function does not have a removable discontinuity. 

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