Let’s now give names to the different types of discontinuities we saw earlier, and look at removable discontinuities.

Recall that we saw on the preceding screen three examples of discontinuities. Each is an example of a different type:

- The top two figures below each shows a
**removable discontinuity**: the graph has a single hole in it at $x =a$, and so we can remove its discontinuity by redefining the function so that $f(a)$ fills in the hole. We’ll discuss this in greater detail below.

- The bottom-left figure shows an
**infinite discontinuity**: $\displaystyle{\lim_{x \to 2^-} = -\infty}$ and $\displaystyle{\lim_{x \to 2^+} = \infty.}$ Any function that has $\displaystyle{\lim_{x \to a^-} = \pm\infty}$ and/or $\displaystyle{\lim_{x \to a^+} = \pm\infty}$ has an infinite discontinuity. - The bottom-right figure shows a
**step**or**gap continuity**: the function has a finite difference in the limit as $x \to a$ from opposite sides.

There isn’t much to say about infinite discontinuities or step (or gap) continuities, so we’re going to focus on removable discontinuities here. (They’re also the ones that usually appear on exams.)

You’ve actually already been working with removable discontinuities, and have even already “removed the discontinuity” many times now; we just didn’t yet have the concept or term “discontinuity” to describe the process we used. The following Example will seem familiar and demonstrates what we mean.

The function $\displaystyle{\lim_{x \to 2} \dfrac{x^2-4}{x-2}}$ is undefined and has a removable discontinuity at $x=2.$

Let’s redefine the function to be

\[f(x) =

\begin{cases}

\dfrac{x^2-4}{x-2}& \text{for } x \ne 2 \\[8px]
c & \text{for } x =2 \\

\end{cases} \]
Find the value of *c* that makes the function continuous.

*Solution.*

Recall that for *f* to be continuous at $x=2,$ we must have

\[\lim_{x \to 2}f(x) = f(2)\]
We thus essentially solved this problem as our very first example of

“Factor to find a limit“, just using different language: then the question was, “Find $\displaystyle{\lim_{x \to 2} \dfrac{x^2-4}{x-2}}.$”

You’ve probably done enough of those types of limit problems by know to be able to complete the steps as well as we can: first factor the function’s numerator, and then reduce the fraction, to find

\[\frac{x^2-4}{x-2} = x+2 \quad \text{for } x \ne 2 \]
Recall that we wrote after that first “Factoring” example:

As you will see in many, *many* problems to come, we will use this basic technique often: through algebraic manipulation, we are able to rewrite a function into a different form which *is* defined at the point of interest. Simple substitution then works to know the value of the function at that point, which in turn is the limit of the function at that point.

There is an important subtlety here: the two functions in the graphs above are *not* the same because they have different domains: the original function, $\dfrac{x^2-4}{x-2},$ is not defined at $x=2,$ whereas our rewritten function is. At every point other than $x=2,$ they are identical, but because they behave differently at this one point they are not identical functions. We can make them identical, however, by simply excluding this one point:

\[\frac{x^2-4}{x-2} = x+2 \quad \text{for } x \ne 2 \]
And because we explicitly don’t care about what’s happening *at* $x=2,$ and instead only what happens *close* to $x=2,$ we have

\[\lim_{x \to 2}\frac{x^2-4}{x-2} =\lim_{x \to 2}(x+2) = 4\]
This is almost the only time we’ll discuss this subtlety, but you should keep in mind that the algebraic manipulations you’ll do again and again allow us to find the limit of the original function only because of the way the limit is defined to focus on the behavior of the function close to, rather than *at*, the point of interest.

Hence earlier, when we were focused only on the limit, we weren’t changing the function itself and so $f(2)$ remained undefined. By contrast, *this* question is explicitly asking us to modify the function by assigning $f(2)$ the correct value so as to make *f* continuous:

\[ f(2) = c = \lim_{x \to 2}f(x) = 4 \quad \cmark\]
Notice that this means we now have two ways of writing the function:

\begin{array}{l}

f(x) =

\begin{cases}

\dfrac{x^2-4}{x-2}& \text{for } x \ne 2 \\[8px]
4 & \text{for } x =2 \\

\end{cases}

&&

\text{and}\quad f(x) = x+2

\end{array}

The two expressions are now entirely equivalent.

You should be prepared for an exam question similar to that of Example 1. Since you’re now so good at finding limits, you know what to do: find the limit at the point of interest $x=a,$ and then simply set $\displaystyle{f(a) = \lim_{x \to a}f(x)}.$

The problems below provide practice. Note that most are from actual previous university-level exams, and we note as we did on the preceding screen that you should expect such problems on your exams.

Practice Problem #1

If
\[f(x) =
\begin{cases}
\dfrac{x^3-2x^{2} + 5x}{x} & \text{for } x \ne 0 \\
C & \text{for } x=0
\end{cases} \]
and if *f* is continuous at $x=0,$ then *C* =
\begin{array}{lllll} \text{(A) }0 && \text{(B) }25 && \text{(C) }5 && \text{(D) }100 && \text{(E) }50 \end{array}

Practice Problem #2

If
\[f(x) =
\begin{cases}
\dfrac{\sqrt{2x+2}-\sqrt{x+3}}{x-1} & \text{for } x \ne 1 \\
k & \text{for } x=1
\end{cases} \]
and if *f* is continuous at $x=1,$ then *k* =
\begin{array}{lllll} \text{(A) }0 && \text{(B) }\dfrac{1}{4} && \text{(C) }\dfrac{1}{2} && \text{(D) }1 && \text{(E)
}\dfrac{3}{2} \end{array}

Practice Problem #3

Consider the function *f* shown. Which of the following statements are true?

I. *f* has a removable discontinuity at $x=-4.$

II.*f* has a removable discontinuity at $x=0.$

III.*f* has a jump discontinuity at $x=0.$

IV.*f* has an infinite discontinuity at $x=4.$

\begin{array}{lllll} \text{(A) I only} && \text{(B) I and II only} && \text{(C) I and III only} && \text{(D) IV only} && \text{(E) I, II and IV only} \end{array}
\begin{array}{llll} \text{(F) I, III and IV} && \text{(G) I and IV only} && \text{(H) II and IV only } && \text{(I) III and IV only} \end{array}II.

III.

IV.

Practice Problem #4

If
\[f(x) =
\begin{cases}
\dfrac{\sin x}{x} & \text{for } x \ne 0 \\
c & \text{for } x= 0
\end{cases} \]
and if *f* is continuous at $x=0,$ then *c* =
\begin{array}{lllll} \text{(A) }0 && \text{(B) }1 && \text{(C) }\pi && \text{(D) }\dfrac{1}{2} && \text{(E) none of these} \end{array}

On the next screen we’ll discuss various continuous functions.

For now, what questions or thoughts do you have about discontinuities? Or are you working on a homework problem where you have to make a function continuous? Please join us over on the Forum to discuss!

- The three types of discontinuities are (1) “removable,” (2) “infinite,” and (3) “jump,” or “gap.”
- If a function has a removable discontinuity at $x=a,$ you can make the function continuous by (re)defining the disconnected point such that $\displaystyle{f(a) = \lim_{x \to a}f(a) }.$

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