Find \(\lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin x}\right)\).

Solution Since \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{x} =\infty\) and \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}=\infty\), then \(\dfrac{1}{x}-\dfrac{1}{\sin\;x}\) is an indeterminate form at \(0^{+}\) of the type \(\infty -\infty.\) We rewrite the difference as a single fraction. \[ \lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin\;x}\right) =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x-x}{x\sin\;x} \]

Then, \(\dfrac{\sin\;x-x}{x\sin\;x}\) is an indeterminate form at \(0^{+}\) of the type \(\dfrac{0}{0}\). Now we can use L’Hôpital’s Rule. \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin\;x}\right) =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x-x}{x\sin\;x}=\lim\limits_{x\rightarrow 0^{+}} \dfrac{\dfrac{d}{\textit{dx}}\left(\sin\;x-x\right) }{\dfrac{d}{\textit{dx}}\left(x\;\sin x\right)}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cos\;x-1}{x\cos\;x+\sin\;x}\\[-20.7pt] &&\qquad\hspace{14.5pc}\color{#0066A7}{\underset{{\hbox{L'H$\hat{\rm o}$pital's Rule}}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ &&\hspace{10pc}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}} \left(\cos\;x-1\right) }{\dfrac{d}{\textit{dx}}\left(x\;\cos\;x+\;\sin\;x\right) } =\lim\limits_{x\rightarrow 0^{+}}\dfrac{-\!\sin\;x}{\left(-x\;\sin\;x+\cos\; x\right) +\cos\;x}\\[-20.8pt] &&\hspace{3.5pc}\color{#0066A7}{\underset{\hbox{Type $\dfrac{0}{0}$; use L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\[-6pt] &&\hspace{10pc}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x}{x\sin\;x-2\cos\;x}=\dfrac{0}{-2}=0 \end{eqnarray*} \]