Using L’Hôpital’s Rule to Find a Limit

Find \(\lim\limits_{x\rightarrow 0}\dfrac{\tan x-\!\sin x}{x^{2}\tan x}\).

Solution \(\dfrac{\tan x-\!\sin x}{x^{2}\tan x}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}\). We simplify the expression before using L’Hôpital’s Rule. Then it is easier to find the limit. \[ \begin{eqnarray*} \dfrac{\tan x-\!\sin x}{x^{2}\tan x}&=&\dfrac{\dfrac{\sin x}{\cos x}-\!\sin x}{x^{2}\cdot \dfrac{\sin x}{\cos x}}=\dfrac{\dfrac{\sin x-\!\sin x\cos x}{\cos x}}{\dfrac{x^{2}\sin x}{\cos x}}=\dfrac{\sin x\left( 1-\cos x\right) }{x^{2}\sin x}=\dfrac{1-\cos x}{x^{2}} \end{eqnarray*} \]

Since \(\dfrac{1-\cos x}{x^{2}}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0},\) we use L’Hôpital’s Rule.

\[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\tan x-\!\sin x}{x^{2}\tan x} =\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ \dfrac{d}{\textit{dx}}\left( 1-\cos x\right) }{\dfrac{d}{\textit{dx}}x^{2}}=\lim\limits_{x \rightarrow 0}\dfrac{\sin x}{2x}\\[-20.7pt] &&\hspace{16.7pc}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}}{{\uparrow }}}\\[6pt] &&\hspace{9.7pc}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=\dfrac{1}{2}\qquad \color{#0066A7}{{\hbox{$\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x} =1$}} } \end{eqnarray*} \]