We are almost at the end of the unit! One of the questions you may have is why we call some discontinuities removable. There are many types of discontinuities as you have explored in previous units, and in this unit, we will focus on removing a discontinuity if it is removable. 

What does removing a discontinuity mean?

Like the name suggests, it means you define the function slightly differently, so that the discontinuous point no longer becomes a problem. Moreover, in a removable discontinuity, the limit exists and you are left with a function that has a “hole” at the point of removable discontinuity. Keep in mind that we’re not changing the actual function, just trying to make the function continuous over the entire graph.

Example 1:

Remove the discontinuity from the function $\frac{\left(x^3-8\right)}{x-2}$ and find the $\underset{x\to 2}{\lim }$ of the function.

First we will have to find the point of discontinuity, and we can do this by first seeing at what point the function doesn’t have a value. This occurs at $x=2$, and now we can expand the numerator to give us $\left(x-2\right)\left(x^2+2x+4\right)$. We see that the $x-2$ can be cancelled from the numerator and denominator, leaving us with $\left(x^2+2x+4\right)$. If we take the limit as x approaches 2 for this, we see that we get a value of 12. So, $\underset{x\to 2}{\lim }\left(x^2+2x+4\right)$ is 12. 

For piecewise functions, we can check the continuity by comparing the left and right side limits of the piecewise function to see if they are equivalent. If the left and right side limits are equal, the limit for the piecewise function exists at that point, but if they are not equivalent, the limit does not exist at that point, and the function isn’t continuous.

Example 2:

Remove the discontinuity from the function $f(x)=\{\begin{array}{ll}6x-7& \text{if }x<5\\ 4x^3-21& \text{if }x>5\end{array}$ if possible. 

Well we know that polynomial functions are continuous, so the only $x$-value we will focus on is $x=5$. 

$\underset{x\to 5^{-}}{\lim }f(x)=23$

$\underset{x\to 5^{+}}{\lim }f(x)=479$

$23\ne 479$, so this is a jump discontinuity, not a removable discontinuity. For a piecewise function to be continuous it has to follow the same definition as all the other functions where the functions on each domain restriction have to approach the same value for the limit to exist.

No matter what value we could give a function at $x=5$ the new function will still be discontinuous unless the old function is changed. 

Now it is time for some practice questions to see how much you’ve learned.

Practice