D.2 Discontinuity types; removable discontinuities

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

Types of 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.

Examples of the three types of discontinuities. First figure, top left: Removable discontinuity: graph shows a straight line with a hole in it. Second figure, top right: Also removable discontinuity: same line with the same hole at x=2, but now with the addition of the lone point (2, 8) that does not lie on the line. Third figure, bottom left: Infinite discontinuity: the function blows up to infinity at x=2, going to negative infinity to the left of x=2, and to positive infinity to the right of x=2. Fourth figure, bottom right: the function's value is -4 for x less than 2, and then suddenly has the value 5 for x greater than or equal to 2.

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.)

Removable discontinuities

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.

Example 1: Remove the discontinuity from $\displaystyle{\lim_{x \to 2} \dfrac{x^2-4}{x-2}}$

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 Problems: Removable Discontinuities

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}

Show/Hide Solution

In order for f to be continuous at $x=0,$ we must have \[f(0) = C = \lim_{x \to 0}f(x)\] So let’s find that limit: \begin{align*} \lim_{x \to 0}\frac{x^3-2x^{2} + 5x}{x} &= \lim_{x \to 0}\frac{x\left(x^2-2x + 5\right)}{x} \\[8px] &= \lim_{x \to 0}\left(x^2-2x + 5\right) \\[8px] &= 0 – 0 + 5 = 5 \quad \blacktriangleleft \end{align*} Hence \[f(0) = C = \lim_{x \to 0}f(x) = 5 \implies \quad\text{ (C)} \quad \cmark\]

Graph of the original function showing the hole at (0, 5)

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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}

Show/Hide Solution

In order for f to be continuous at $x=1,$ we must have \[f(1) = k = \lim_{x \to 1}f(x)\] So let’s find that limit: \begin{align*} \lim_{x \to 1}\frac{\sqrt{2x+2}-\sqrt{x+3}}{x-1} &= \lim_{x \to 1}\frac{\sqrt{2x+2}-\sqrt{x+3}}{x-1} \cdot \frac{\sqrt{2x+2}+\sqrt{x+3}}{\sqrt{2x+2}+\sqrt{x+3}} \\[8px] &= \lim_{x \to 1} \frac{\sqrt{2x+2}\sqrt{2x+2} + \cancel{\sqrt{2x+2}\sqrt{x+3}} – \cancel{\sqrt{x+3}\sqrt{2x+2}} -\sqrt{x+3}\sqrt{x+3} }{(x-1) \left[ \sqrt{2x+2}+\sqrt{x+3}\right]} \\[8px] &= \lim_{x \to 1} \frac{(2x+2) – (x+3)}{(x-1) \left[ \sqrt{2x+2}+\sqrt{x+3}\right]} \\[8px] &= \lim_{x \to 1} \frac{x -1}{(x-1) \left[ \sqrt{2x+2}+\sqrt{x+3}\right]} \\[8px] &= \lim_{x \to 1} \frac{1}{ \sqrt{2x+2}+\sqrt{x+3}} \\[8px] &= \frac{1}{\sqrt{2+2}+ \sqrt{1+3}} = \frac{1}{2+2} \\[8px] &= \frac{1}{4} \quad \blacktriangleleft \end{align*} Then finally we must have \[k =\lim_{x \to 1}\frac{\sqrt{2x+2}-\sqrt{x+3}}{x-1} = \frac{1}{4} \implies \quad\text{ (B)} \quad \cmark \]

Graph of function showing the hole at (1, 0.25)

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Practice Problem #3

Graph of the function. Line segment tarts with a closed dot (at -6, -2.3) and goes upward toward an open circle at (0, -1). There is also an open circle on the line segment at (-4, -2). Then a curve that look something like a cubic, coming down from infinity just to the right of the y-axis, passing through (2,0), and then headed to negative infinity as it approaches x=4. Then a similar looking curve that comes down from infinity to the right of x=4, passes through (5, 0), and the ends on a closed dot at (6.2, -2).

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}

Show/Hide Solution

I. f has a removable discontinuity at $x=-4$: TRUE ✓
Redefining the single point of the function such that $f(-4) = -2$ would remove the discontinuity.

II. f has a removable discontinuity at $x=0$: FALSE ✗
Even if we “filled in the hole” at $x=0,$ the function would still have an infinite discontinuity from the right. Hence this is not a removable discontinuity.

III. f has a jump discontinuity at $x=0$: FALSE ✗
Since $\displaystyle{\lim_{x \to 0^+}=\infty,}$ there is not a finite difference between the limit from the left and from the right at $x=0.$

IV. f has an infinite discontinuity at $x=4$: TRUE ✓
$\displaystyle{\lim_{x \to 4^-}=-\infty}$ while $\displaystyle{\lim_{x \to 4^+}=\infty,}$ which is one of the ways an infinite discontinuity can occur.

Hence the only true statements are I and IV $\implies \;\text{ (G)} \quad \cmark$

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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}

Show/Hide Solution

For f to be continuous at $x=0,$ we must have \[f(0) = c = \lim_{x \to 0}f(x)\] Early in the semester you just have to remember the “special limit”: \[\lim_{x \to 0}\frac{\sin x}{x} = 1 \quad \blacktriangleleft\] Then for f to be continuous at $x=0,$ we must have \[f(0) = c = \lim_{x \to 0}\frac{\sin x}{x} = 1 \implies \quad\text{ (B)} \quad \cmark\]

Graph of sin(x) over x, showing the limit equals 1 at x=0

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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 Upshot

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 $\displaystyle{f(a) = \lim_{x \to a}f(a) }.$