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

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.*

Graph of $f(x) = \dfrac{1}{(x-2)^2}$ versus *x*,

showing the vertical asymptote at $x=2$

showing the vertical asymptote at $x=2$

Vertical asymptote

In general, the vertical line $x=a$ is called a The line $x = a$ is a vertical asymptote of the function $f(x)$ if at least one of the following statements is true:

1. $\displaystyle{\lim_{x \to a^-}f(x) = -\infty \; \text{ or }\;\infty }$

2. $\displaystyle{\lim_{x \to a^+}f(x) = -\infty \; \text{ or }\;\infty }$

2. $\displaystyle{\lim_{x \to a^+}f(x) = -\infty \; \text{ or }\;\infty }$

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.

A rational function has vertical asymptotes where the denominator is zero and the numerator is not.

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.)

Graph of $s(x) = \tan(x)$ versus *x*,

showing the vertical asymptotes

showing the vertical asymptotes

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 $x \to 0^+$

Recall that $\ln(x)$ is the inverse function of $y = e^x.$ By choosing

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Graph of $k(x) = \ln(x)$ versus *x* showing the vertical asymptote at $x=0$

- A function has a vertical asymptote at $x=a$ if the graph of the function grows without bound in either the positive or negative direction as $xs \to a$ from either the left or the right, or from both directions.

This screen concludes our introduction to limits. In the next Section, we’ll help you develop the skills you need to compute limits quickly using a variety of techniques. (Finally you’ll see how all of those algebraic techniques like factoring and rationalizing you’ve learned are super-useful!)

Questions or comments about anything on this screen? Please pop over to the Forum and post!

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}

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}

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}

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}

\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}

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