Discontinuities and Asymptotes – Calculus Problem of the Week

This Problem of the Week is requires finding the discontinuities and vertical and horizontal asymptotes of a rational function. It’s based on several typical exam questions of this sort. Let’s get right to it!

Solution.

Note: There’s a graph of the function at the bottom of this solution.

Each Problem of the Week is meant to check your understanding as practice for an exam, and so we don’t go into a lot of detail. We’ll instead be drawing directly on the ideas we developed in our main materials on continuity and vertical and horizontal asymptotes. If you’d like to learn more, please visit those screens, where there’s a lot more information and practice problems.

(a) To determine where f is discontinuous, first note that the numerator, $2x -2,$ is continuous everywhere, so we can disregard that for the moment. Focusing on the denominator, let’s factor to identify where the function is undefined: \begin{align*} f(x) &= \frac{2x – 2}{x^2 + x -2} \\[8px] &= \frac{2x-2}{(x+2)(x-1)} \end{align*} So the function is discontinuous where it is undefined, at x = $-2$ and x = 1. $\; \cmark$

(b) Let’s find the limit at x = $-2$ and x = 1:

At $x = -2$:

\begin{align*} \lim_{x \to -2}\frac{2x – 2}{x^2 + x -2} &= \lim_{x \to -2}\frac{2\cancel{(x-1)}}{(x+2)\cancel{(x-1)}} \\[8px] &= \lim_{x \to -2}\frac{2}{x+2} = \text{Does Not Exist (DNE)} \quad \cmark \end{align*}

That is the correct answer, and on an exam we suggest that you stop there as your final answer. (No reason to write more than requested, especially if you might have a slip of the pencil and write something incorrect that lowers your score.) For our purposes here, let’s note that

\[\lim_{x \to -2^+}\frac{2}{x+2} = \infty \quad \text{and} \quad \lim_{x \to -2^-}\frac{2}{x+2} = -\infty\]

The last statement merely provides more information about the way in which the limit does not exist. We’ll make use of this when we graph the function below.

Now, at $x=1$:

\begin{align*} \lim_{x \to 1}\frac{2x – 2}{x^2 + x -2} &= \lim_{x \to 1}\frac{2(x-1)}{(x+2)(x-1)} \\[8px] &= \lim_{x \to 1}\frac{2}{x+2} \\[8px] &= \frac{2}{3} \quad \cmark \end{align*}

Although the question doesn’t ask, the function thus has a removable discontinuity at $x = 1.$

(c) We now consider the asymptotes.

We already found a vertical asymptote in part (b), since $\displaystyle{\lim_{x \to -2^+}\frac{2}{x+2} = \infty}$ and $\displaystyle{\lim_{x \to -2^-}\frac{2}{x+2} = -\infty.}$ Hence

$x = -2$ is a vertical asymptote $\; \cmark$

To find the horizontal asymptotes, we consider the limit as $x \to \pm \infty$:

\[\lim_{x \to -\infty}\frac{2}{x+2} = 0 \quad \text{and} \quad \lim_{x \to \infty}\frac{2}{x+2} = 0 \]

Hence

y = 0 is a horizontal asymptote $\quad \cmark$

(d) We need to determine a and b such that $g(x) = f(x)$ everywhere f is defined:

\begin{align*} \frac{a}{b+x} &= \frac{2x – 2}{x^2 + x -2} \\[8px] &= \frac{2}{x+2} \end{align*}

Since the numerators of both functions must equal each other, and the denominators must equal each other, we have

a = 2,   and   b = 2 $\; \cmark$

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