Activity 41.
Evaluate each of the following limits. If you use L鈥橦么pital鈥檚 Rule, indicate where it was used, and be certain its hypotheses are met before you apply it.
(a)
\(\displaystyle \lim_{x \to \infty} \frac{x}{\ln(x)}\)
Hint.
Remember that \(\ln(x) \to \infty\) as \(x \to \infty\text{.}\)
Answer.
\(\lim_{x \to \infty} \frac{x}{\ln(x)} = \infty\text{.}\)
Solution.
As both numerator and denominator tend to \(\infty\) as \(x \to \infty\text{,}\) by L鈥橦么pital鈥檚 Rule followed by some elementary algebra,
\begin{equation*}
\lim_{x \to \infty} \frac{x}{\ln(x)} = \lim_{x \to \infty} \frac{1}{\frac{1}{x}} = \lim_{x \to \infty} x = \infty\text{.}
\end{equation*}
(b)
\(\displaystyle \lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2}\)
Hint.
Both the numerator and denominator tend to \(\infty\) as \(x \to \infty\text{.}\)
Answer.
\(\lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \frac{1}{2}\text{.}\)
Solution.
Because this limit has indeterminate form \(\frac{\infty}{\infty}\text{,}\) L鈥橦么pital鈥檚 Rule tells us that
\begin{equation*}
\lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \lim_{x \to \infty} \frac{e^{x} + 1}{2e^{x} + 2x}\text{.}
\end{equation*}
The latest limit is indeterminate for the same reason, and a second application of the rule shows
\begin{equation*}
\lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \lim_{x \to \infty} \frac{e^{x}}{2e^{x} + 2}\text{.}
\end{equation*}
Note how each application of the rule produces a simpler numerator and denominator. With one more use of L鈥橦么pital鈥檚 Rule, followed by a simple algebraic simplification, we have
\begin{equation*}
\lim_{x \to \infty} \frac{e^{x} + x}{2e^{x} + x^2} = \lim_{x \to \infty} \frac{e^{x}}{2e^{x}} = \lim_{x \to \infty} \frac{1}{2} = \frac{1}{2}\text{.}
\end{equation*}
(c)
\(\displaystyle \lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}}\)
Hint.
Note that \(x \to 0^+\text{,}\) not \(\infty\text{.}\)
Answer.
\(\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = 0\text{.}\)
Solution.
As \(x \to 0^+\text{,}\) \(\ln(x) \to -\infty\) and \(\frac{1}{x} \to +\infty\text{,}\) thus by L鈥橦么pital鈥檚 Rule,
\begin{equation*}
\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}\text{.}
\end{equation*}
Reciprocating, multiplying, and simplifying, it follows that
\begin{equation*}
\lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{1}{x}\cdot \frac{x^2}{-1} = \lim_{x \to 0^+} -x = 0\text{.}
\end{equation*}
(d)
\(\displaystyle \lim_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}}\)
Hint.
As \(x \to \frac{\pi}{2}^-\text{,}\) \(\tan(x) \to \infty\text{.}\)
Answer.
\(\lim_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}} = -\infty\text{.}\)
Solution.
Here, the numerator tends to \(\infty\) while the denominator tends to \(0^-\text{.}\) Note well that this limit is not indeterminate, but rather produces a collection of fractions with large positive numerators and small negative denominators. Hence
\begin{equation*}
\lim_{x \to \frac{\pi}{2}^-} \frac{\tan(x)}{x-\frac{\pi}{2}} = -\infty\text{.}
\end{equation*}
In particular, we observe that L鈥橦么pital鈥檚 Rule is not applicable here.
(e)
\(\displaystyle \lim_{x \to \infty} xe^{-x}\)
Hint.
Observe that \(e^{-x} = \frac{1}{e^x}\text{.}\)
Answer.
\(\lim_{x \to \infty} xe^{-x} = 0\text{.}\)
Solution.
In its original form, \(\lim_{x \to \infty} xe^{-x}\text{,}\) is indeterminate of form \(\infty \cdot 0\text{.}\) Rewriting \(e^{-x}\) as \(\frac{1}{e^x}\text{,}\) a straightforward application of L鈥橦么pital鈥檚 Rule tells us that
\begin{equation*}
\lim_{x \to \infty} xe^{-x} = \lim_{x \to \infty} \frac{x}{e^x} = \lim_{x \to \infty} \frac{1}{e^x}\text{.}
\end{equation*}
Since \(e^x \to \infty\) as \(x \to \infty\text{,}\) we find that
\begin{equation*}
\lim_{x \to \infty} xe^{-x} = 0\text{.}
\end{equation*}
