Section 12.3 Vertical Asymptotes: “Infinite” Limits
It should be clear from the sketch below that in addition to the horizontal asymptote \(y=0\text{,}\) the graph of \(y=\frac1x\) also has a vertical asymptote of \(x=0\text{.}\)
We can extend the ideas and the notation we’ve developed for horizontal asymptotes to handle vertical asymptotes as well. First note that since the expression \(\frac1x\) is not defined for \(x=0\) we can’t talk about the behavior of the function at zero. But we can talk about its behavior near zero, and notice the graph is very different depending on whether \(x\lt0\) or \(x\gt0\text{.}\) We will focus first on positive values of \(x\text{.}\)
For \(x\) near zero, and positive, it is very clear what is happening. As \(x\rightarrow0\) the function continues to increase without bound. We capture this notationally with
\begin{equation}
\limitX{x}{0}{\frac1x}{x\gt0}=\infty.\tag{12.1}
\end{equation}
The limit in
equation (12.1) is called a
right–hand limit and is read aloud as “the limit as
\(x\) approaches zero from the right, of one over
\(x\) equals infinity.”
The expression \(x\rightarrow0\) means that \(x\) is approaching the number zero. The qualification \(x\gt0\) specifies that it is approaching from the right. So writing\(\limitX{x}{0}{\frac1x}{x\gt0}=\infty\) means that as \(x\rightarrow0\) from the right, the function \(\frac1x\) increases without bound.
If we restrict \(x\) to negative values then we see that
\begin{equation}
\limitX{x}{0}{\frac1x}{x\lt 0}=-\infty.\tag{12.2}
\end{equation}
Note that the sense of the inequality has changed from
equation (12.1). It is now
\(x\lt
0\text{.}\)
The limit in
equation (12.2) is called a
left–hand limit and is read aloud as “the limit as
\(x\) approaches zero from the left, of one over
\(x\) equals negative infinity.” Right- and left-hand limits are generically called
one-sided limits. One-sided limits are important theoretical tools but their theoretical value is not important to us right now. We will discuss them in more depth in
Section 16.2.
A function \(f(x)\text{,}\) will have a vertical asymptote as \(x\rightarrow a\) if either of the one-sided limits is \(\infty\) or \(-\infty\) (\(f(x)\) is increasing or decreasing without bound). We formalize this statement in the following definition.
Definition 12.3.1. Vertical Asymptotes.
If
\begin{align*}
\rlimit{x}{a}{f(x)}=\pm\infty\amp{}\amp{} \text{ or }\amp{}\amp{}\llimit{x}{a}{f(x)}=\pm\infty
\end{align*}
then the line \(x=a\) is a vertical asymptote of the graph of \(y=f(x)\text{.}\)
Drill 12.3.2.
Compute each of the following limits. Read the notation carefully.
\(\displaystyle \rtlimit{x}{3}{\frac{1}{(x-1)(x-3)}}\)
\(\displaystyle \ltlimit{x}{3}{\frac{1}{(x-1)(x-3)}}\)
\(\displaystyle \ltlimit{x}{1}{\frac{1}{(x-1)(x-3)}}\)
\(\displaystyle \rtlimit{x}{1}{\frac{1}{(x-1)(x-3)}}\)
\(\displaystyle \rtlimit{x}{\frac{\pi}{2}}{\tan(x)}\)
\(\displaystyle \ltlimit{x}{\frac{\pi}{2}}{\tan(x)}\)
\(\displaystyle \ltlimit{x}{\frac{\pi}{2}}{\sec(x)}\)
\(\displaystyle \rtlimit{x}{\frac{\pi}{2}}{\sec(x)}\)
Problem 12.3.3.
Find all vertical asymptotes of the graphs of each of the following functions.
(a)
\(f(x)= \frac{1}{x^2-1}\)
(b)
\(f(x)= \frac{1}{(x^2-1)^2}\)
(c)
\(f(x)= \frac{x-1}{x^2-4}\)
(d)
\(f(x)= \frac{x-7}{x^2-5}\)
(e)
\(f(x)= \tan(x)\)
(f)
\(f(x)= \sec(x)\)
Notice that so far in this section we have only discussed one-sided limits like
\begin{align*}
\llimit{x}{a}{f(x)} =\pm\infty \amp{}\amp{} \text{and}
\amp{}\amp{}\rlimit{x}{a}{f(x)}=\pm\infty,
\end{align*}
but we have carefully avoided assigning any meaning to a formula like
\begin{align}
\limit{x}{a}{f(x)}=\infty \amp{}\amp{} \text{or} \amp{}\amp{}
\limit{x}{a}{f(x)}=-\infty.\tag{12.3}
\end{align}
This is not because the topic is inherently difficult. We’re just not quite ready to discuss it yet.
In fact, at this point you can probably guess the meanings of the expressions in
formula (12.3) without any help from us.
Drill 12.3.4.
Write down as clearly as you can what you believe the expressions
\(\limit{x}{a}{f(x)}=\infty \) and
\(\limit{x}{a}{f(x)}=-\infty \) mean and hold on to your answer. We will return to this topic in
Drill 17.2.35. When we get there you can compare your (educated) guess with our definition.
