Find:

Solution (a) Since $\lim\limits_{x\rightarrow \infty }\ln x=\infty$ and $\lim\limits_{x\rightarrow \infty }x=\infty$, $\dfrac{\ln x}{x}$ is an indeterminate form at $\infty$ of the type $\dfrac{\infty }{\infty }$. Using L’Hôpital’s Rule, we have $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{\ln x}{x} = \lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}\ln x}{\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{1}{x}}{1}=\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}=0\\[-1.8pc] &&\hspace{3.35pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*}$$

(b) $\lim\limits_{x\rightarrow \infty }x=\infty$ and $\lim\limits_{x\rightarrow \infty }e^{x}=\infty$, so $\dfrac{x}{e^{x}}$ is an indeterminate form at $\infty$ of the type $\dfrac{\infty }{\infty }$. Using L’Hôpital’s Rule, we have $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{x}{e^{x}}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}x}{\dfrac{d}{\textit{dx}}e^{x}}=\lim\limits_{x\rightarrow \infty } \dfrac{1}{e^{x}}=0\\[-1.8pc] &&\hspace{2.5pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*}$$

(c) From (b), we know that $\dfrac{e^{x}}{x}$ is an indeterminate form at $\infty$ of the type $\dfrac{\infty }{\infty }$. Using L’Hô pital’s Rule, we have $$\begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}e^{x}}{\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow \infty } \dfrac{e^{x}}{1}=\infty \\[-1.8pc] &&\hspace{2.5pc}\color{#0066A7}{\underset{ \begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array} }{{\uparrow }}} \end{eqnarray*}$$