What is the difference between a removable and nonremovable discontinuity?

Hint: In the given problem, we are trying to find the differences between a removable and non-removable discontinuity. To start with, we will try to get more information about the different types of continuity. Finding the points, we get the difference among them and find our results.

Complete step by step solution:
According to the problem, we are trying to find the differences between a removable and non-removable discontinuity.
Talking of a removable discontinuity, it is a hole in a graph. That is, a discontinuity that can be “repaired” by filling in a single point. In other words, a removable discontinuity is a point at which a graph is not connected but can be made connected by filling in a single point.
And, now, if we talk about non-removable discontinuity, it is a type of discontinuity in which the limit of the function does not exist at a given particular point, i.e, $\displaystyle \lim_{x \to a}f\left( x \right)$ does not exist. We can simply say that the value of $f\left( a \right)$ at the function with x = a(which is the point of discontinuity) may or may not exist but the $\displaystyle \lim_{x \to a}f\left( x \right)$ does not exist. This non-removable type of discontinuities can be further divided into three types which are finite type, infinite type and oscillatory type of discontinuities.
Getting the points altogether, Geometrically, a removable discontinuity is a hole in the graph of f.
A non-removable discontinuity is any other kind of discontinuity. (Often jump or infinite discontinuities.)

Note: The graph of a removable discontinuity leaves you feeling empty, whereas a graph of a non removable discontinuity leaves you feeling jumpy . If a term doesn't cancel ,the discontinuity at this X value corresponding to this term for which the denominator is zero is non-removable, and the graph has a vertical asymptote.