Question

How can you remove a discontinuity ?

Answer 1

In order to remove a discontinuity first check whether the discontinuity is a type of removable discontinuity or non-removable discontinuity. If the discontinuity is removable then only the discontinuity can be removed. Removable discontinuities are also called holes which can be filled or removed by redefining the function.

Complete step by step answer:
Since, we know that there are two types of discontinuities that are removable and non-removable. A discontinuity can be removed if the discontinuity is a type of removable discontinuity. The discontinuity can be removed by re-defining the function value for filling the holes which is leading to discontinuity.

A discontinuity at $x = a$ is said to be removable if $\mathop {\lim }\limits_{x \to a} f\left( x \right)$ exists, we call it $L$. But $L \ne f\left( a \right)$ this is because either $f\left( a \right)$ is some number other than $L$ or $f\left( a \right)$ is not defined.
We can remove this discontinuity by redefining the function, let’s call it $g\left( x \right)$
Now it is

& f(x),x\ne a \\\ & L, x=a \\\ \end{aligned} \right.$$ which gives $g\left( x \right) = f\left( x \right)$ for all $x \ne a$ and $g\left( x \right)$ is continuous at $a$ and in this way the hole is filled. Let’s understand it more deeply by an example. We are given with a function, $f\left( x \right) = \dfrac{{{x^2} - 1}}{{x - 1}}$which is discontinuous at $x = 1$,$f\left( 1 \right)$ does not exist. According to the question of the example the function is discontinuous at $1$ , But if we check $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - 1}}{{x - 1}}$ On further solving it we can write ${x^2} - 1 = \left( {x - 1} \right)\left( {x + 1} \right)$ Now, $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)}}$ Cancelling we have $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \left( {x + 1} \right)$ Now applying the limit we have $f(1) = 1 + 1 = 2$ Which gives $f\left( 1 \right) = 1 + 1 = 2$ So, by redefining the function to remove this continuity and we get: $$g(x)=\left\\{ \begin{aligned} & \dfrac{x^2-1}{x-1},x\ne a \\\ & 2, x=1 \\\ \end{aligned} \right.$$ was showing that the function was discontinuous at $1$ but when we filled the gap by adding the point at $x = 1$, it became continuous and the discontinuity was removed. **Note:** Before solving, always check whether the discontinuity is removable or non-removable, it can be removed or not. Choose a point where the hole is to be filled. There can be an error in redefining the function.