Using L’Hôpital’s Rule to Find a One-Sided Limit

Find \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cot x}{\ln x}\).

Solution Since \(\lim\limits_{x\rightarrow 0^{+}}\cot x=\infty \) and \(\lim\limits_{x\rightarrow 0^{+}}\ln x=-\infty \), \(\dfrac{\cot x}{\ln x}\) is an indeterminate form at \(0^{+}\) of the type \(\dfrac{\infty }{\infty }\). Using L’Hôpital’s Rule, we find \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cot x}{\ln x}=\lim\limits_{x\rightarrow 0^{+}} \dfrac{\dfrac{d}{\textit{dx}}\cot x}{\dfrac{d}{\textit{dx}}\ln x}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{-\csc ^{2}x}{\dfrac{1}{x}}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{x}{ \sin ^{2}x}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}x}{\dfrac{d}{\textit{dx}}\sin ^{2}x}\\[-21pt] &&\hspace{4pc}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}} {{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\quad \ \hspace{10.5pc} \color{#0066A7}{\underset{\hbox{$\csc ^{2}x=\dfrac{1}{\sin^{2}x}$}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\quad \, \hspace{4.5pt}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ &&\hspace{7pc}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{2\sin x\cos x}=\left( -\dfrac{1}{2}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\cos x}\right)\\[7pt] &&\hspace{7pc}=\left( -\dfrac{1}{2}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}\right) ( 1) =-\infty \end{eqnarray*} \]