A.7 Vertical Asymptotes

On this screen we examine vertical asymptotes, which result when a function “blows up” and grows toward positive or negative infinity at a particular input value. We’ll use interactive Desmos graphing calculators so you can see how this works for yourself, and of course provide practice problems with complete solutions.

As you know, one way in which a function is undefined for a particular input value $x=a$ is that the function’s output “blows up” near there, meaning its output increases without bound in either the positive (upward) or negative (downward) direction as you approach that x input value. As we saw two screens ago, the limit at $x=a$ then does not exist.

Revisiting $f(x) = \dfrac{1}{(x-2)^2}$

Graph of f(x), showing vertical asymptote at x=2: the function grows and grows and grows in the positive direction as we approach x=2 from either side

We saw this for the function $f(x) = \dfrac{1}{(x-2)^2}$ when examining its behavior near $x=2,$ as shown in the graph to the right.

As we noted earlier, since the function grows without bound as $x \to 2,$ $\displaystyle{\lim_{x \to 2}f(x)}$ does not exist (DNE); instead we can make the function’s output as large as we would like, without bound, by moving sufficiently close to $x=2.$

If we want to convey this particular way in which the limit does not exist, we write
\[\lim_{x \to 2}f(x) = \infty \;\text{ (DNE)}\] since the function’s output grows and Grows and GROWS in the positive (upward) direction as we approach $x=2$ from either the left or from the right. Remember that this statement still means that the limit does not exist, and merely supplies additional information by specifying the way in which the limit does not exist.

Notice in the interactive graph below that the function resembles the vertical line $x=2$ for x near 2, which becomes more and more evident as you zoom out. That line is called a vertical asymptote of the graph of f.

Vertical asymptotes

Rational functions and vertical asymptotes
Problems that ask about vertical asymptotes often deal with rational functions. Recall that a rational function is a fraction that a polynomial in both the numerator and denominator.

The function above, $f(x) = \dfrac{1}{(x-2)^2},$ is an example of such a function: its vertical asymptote is at $x=2,$ where the numerator is not zero but the denominator is, so the function is undefined there. We’ll see other examples of vertical asymptotes in rational functions in the practice problems below.

$\tan(x)$ and vertical asymptotes
As another example of a vertical asymptote, consider the function $g(x) = \tan(x),$ which is graphed in the interactive Desmos calculator below. Recall that $\tan(x)$ is not defined for $x = \pm \dfrac{\pi}{2},$ $\pm \dfrac{3\pi}{2},$ $\pm \dfrac{5\pi}{2},$ …. (That’s because $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$, and so everywhere $\cos(x) = 0,$ $\tan(x)$ is undefined.)

Let’s focus on $x = \dfrac{\pi}{2}.$ Since there is no single output value L we can get as close to as we’d like by being sufficiently close to $x = \dfrac{\pi}{2},$ we know that the limit does not exist:
\[\lim_{x \to \pi/2}\tan(x) = \text{DNE}\] And if we want to convey the way in which the limit does not exist, we can write
\[ \lim_{x \to \pi/2^-}\tan(x) = \infty \;\text{ and } \lim_{x \to \pi/2^+}\tan(x) = -\infty \] (Notice the one-sided limits there: make sure the statements, including the notation with the “–” and “+,” make sense to you. If not, please review one-sided limits.) We can make the function’s output as large and positive as we’d like by being sufficiently close to $x= \dfrac{\pi}{2}$ from the left, and we can make its output as large and negative as we’d like by being sufficiently close to $x= \dfrac{\pi}{2}$ from the right.

Because the function resembles the vertical line $x=\dfrac{\pi}{2}$ for x near $\dfrac{\pi}{2},$ that line is a vertical asymptote for the function. Similarly, so are the vertical lines at $x = -\dfrac{\pi}{2},$ $\pm \dfrac{3\pi}{2},$ $\pm \dfrac{5\pi}{2},$ and so forth.

$\ln(x)$ and vertical asymptotes
As a yet another example, let’s consider the function $k(x) = \ln(x).$ Its interactive graph is below. As you can see, as $x \to 0$ from the right, the function grows in the negative direction without bound. Hence
\[\lim_{x \to 0^+} = \text{DNE}\] Notice in this case we must specify that we are considering only the right-side limit, since the function is only defined for $x \gt 0.$ And if we wish to convey more about the function’s behavior near $x=0,$ we write
\[\lim_{x \to 0+}\ln(x) = -\infty\] meaning we can make the function’s output as large a negative number as we would like by being sufficiently close to $x=0.$

Open for review of why $\ln(x)$ becomes more and more negative as $x \to 0^+$

Graph of both y = e^x and y = ln(x), showing how they are reflected across the line y = x since they are inverse functions. Also the limit statements lim x to negative infinity of e^x = 0, and lim x to 0 of ln(x) = negative infinity.

Recall that $\ln(x)$ is the inverse function of $y = e^x.$ By choosing x to be large and negative in $y = e^x,$ you can make y as close to zero as you’d like . . . but you can never make it exactly zero. This fact determines the behavior of the inverse function $y = \ln(x)$: the closer x is to 0 (with the restriction that $x \gt 0$), the larger and more negative y becomes.

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The Upshot

Practice Problems

The problems below will give you practice working with vertical asymptotes in rational functions, which commonly occur on exams.

Practice Problem #1

The function $f(x) = \dfrac{5}{x-3}$ has a vertical asymptote at $x=$ \begin{array}{lllll} \text{(A) }0 && \text{(B) }5 && \text{(C) }3 && \text{(D) }-3 && \text{(E) none of these} \end{array}

Show/Hide Solution

We have $f(x) = \dfrac{5}{x-3}.$ Its vertical asymptote occurs where the denominator is zero and the numerator is not. For this simple function we can see by inspection that the denominator is zero when \[x = 3 \implies \quad\text{ (C)} \quad \cmark\] The figure shows the vertical asymptote at $x=3.$

Hyperbola centered at x=3, and a vertical red line showing the asymptote x=3.

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Practice Problem #2

The function $f(x) = \dfrac{-1}{x^2-16}$ has vertical asymptote(s) at $x=$ \begin{array}{lllll} \text{(A) }4 && \text{(B) }16 && \text{(C) }-16 && \text{(D) }\pm 4 && \text{(E) -4} \end{array}

Show/Hide Solution

We have $f(x) = \dfrac{-1}{x^2-16}.$ Its vertical asymptotes occur where the denominator is zero and the numerator is not. Hence the asymptotes occur when \begin{align*} x^2 – 16 &= 0 \\[8px] x^2 &= 16 \\[8px] x &= \pm 4 \implies \quad\text{ (D)} \quad \cmark \end{align*}

When taking a square root, remember to retain both the positive and negative resulting values. Otherwise you lose the negative value, rendering your answer incorrect. Here, for instance, you would be missing the vertical asymptote at $x=-4,$ as the graph shows. Before choosing your final answer, ask yourself: what are all of the possible values of x that make the statement $x^2 = 16$ true?

Graph showing the function and the vertical asymptotes at x= -4 and at x = 4

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Practice Problem #3

Find the $x$-value(s) of the vertical asymptote(s) of $f(x)=\dfrac{1}{\left( x^2 + 3 \right)(2x - 1)}$. \begin{array}{lllll} \text{(A) }- \dfrac{1}{2} && \text{(B) }0 && \text{(C) } \dfrac{1}{2} && \text{(D) }- \sqrt{3} && \text{(E) none of the these} \end{array}

Show/Hide Solution

We are given the function, $g(x)=\dfrac{1}{\left( x^2 + 3 \right)\left( 2x – 1 \right)}$. The vertical asymptote(s) of a rational function occurs when the denominator is zero and the numerator is not. Hence we set the denominator equal to zero and solve for x: \[ g(x) = \left( x^2 + 3 \right) \left( 2x – 1 \right) \] Note that the first factor, $\left( x^2 + 3 \right)=0,$ does not have any real solutions. Then turning our attention to the second factor: \begin{align*} (2x-1) &= 0 \\[8px] 2x &= 1 \\[8px] x &= \dfrac{1}{2} \implies \quad \text{ (C) } \quad \cmark \end{align*} The graph shows the single asymptote at $x= \dfrac{1}{2}.$

Graph showing vertical asymptote at x= 1/2

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The following problem switches things up a bit, and asks you to determine a value in the function such that it has a vertical asymptote at a particular x-value.

Practice Problem #4

If $f(x)= \dfrac{2}{2x^2 +2x + c}$, find the value of $c$ for which $f(x)$ has a vertical asymptote at $x=- \dfrac{1}{2}$.
\begin{array}{lllll} \text{(A) }- \dfrac{1}{2}&& \text{(B) }\dfrac{1}{2} && \text{(C) }2 && \text{(D) }-4 && \text{(E) None of these}\end{array}

Show/Hide Solution

The function given is: $f(x)=\dfrac{2}{2x^2+2x+c}$. The vertical asymptote(s) of a rational function occurs when the denominator is zero and the numerator is not. Hence we set the denominator equal to zero and solve for c: \begin{align*} 2x^2+2x+c &= 0 \\[8px] c &= -2x^2 – 2x \end{align*} We want the vertical asymptote at $x = – \dfrac{1}{2},$ so \begin{align*} c &= -2\left( -\frac{1}{2}\right)^2 -2 \left( -\frac{1}{2}\right) \\[8px] c &= -\frac{1}{2} + 1 \\[8px] c &= \dfrac{1}{2} \implies \quad \text{ (B) } \quad \cmark \end{align*} Note that when $c=\dfrac{1}{2}$, the graph exhibits a vertical asymptote at $x=- \dfrac{1}{2}$ as requested.

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