Example 11.1. Vertical Asymptotes and One-Sided Limits.
In this example, we will examine a rational function and use limits to determine its vertical asymptotes.
> f(x) := (x^2-x-6)/(x^2-8*x+15);
\begin{equation*}
\displaystyle f\, := \,x\mapsto {\frac {{x}^{2}-x-6}{{x}^{2}-8\,x+15}}
\end{equation*}
We can factor the denominator to find the domain of \(f(x)\) and predict where we might find vertical asymptotes. To do this, it is helpful to use the
denom() command, which allows us to obtain the denominator of \(f(x)\text{.}\)
> factor( denom( f(x) ) );
\begin{equation*}
\displaystyle \left( x-3 \right) \left( x-5 \right)
\end{equation*}
It looks like \(x=3\) and \(x=5\) are not in the domain of \(f(x)\text{,}\) though it is not clear if they are vertical asymptotes. We can find the limit of \(f(x)\) as \(x \rightarrow 3\text{.}\)
> limit(f(x), x=3);
\begin{equation*}
\displaystyle -5/2
\end{equation*}
Since this limit exists but \(f(3)\) does not, this is a removable discontinuity and not a vertical asymptote. Now we can find the limit of \(f(x)\) as \(x \rightarrow 5\text{.}\)
> limit(f(x), x=5);
\begin{equation*}
\displaystyle \textit{undefined}
\end{equation*}
Even though this limit does not exist, we cannot automatically conclude that \(f(x)\) has a vertical asymptote at \(x=5\text{.}\) We need to compute the one-sided limits to see if there is asymptotic behaviour.
Aside
> limit(f(x), x=5, left);
\begin{equation*}
\displaystyle -\infty
\end{equation*}
> limit(f(x), x=5, right);
\begin{equation*}
\displaystyle \infty
\end{equation*}
Since these limits are given as \(\pm \infty\text{,}\) we know that \(f(x)\) has a vertical asymptote at \(x=5\text{.}\)
