Find \(\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}\).
Step 1 Since \(\lim\limits_{x\rightarrow 0}\left( \sin x-x\right) =0\) and \(\lim\limits_{x\rightarrow 0}x^{2}=0\), the expression \(\dfrac{\sin x-x}{x^{2}}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}.\)
Steps 2 and 3 We use L’Hôpital’s Rule. \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ \dfrac{d}{\textit{dx}}\left( \sin x-x\right) }{\dfrac{d}{\textit{dx}}x^{2}}=\lim\limits_{x \rightarrow 0}\dfrac{\cos x-1}{2x}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0} \dfrac{\cos x-1}{x}\\ &&\hspace{8pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*} \]
Since \(\lim\limits_{x\rightarrow 0}( \cos x-1) =0\) and \(\lim\limits_{x\rightarrow 0}x=0\), the expression \(\dfrac{\cos x-1}{x}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}\). So, we use L’Hôpital’s Rule again. \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}=\dfrac{1}{2} \lim\limits_{x\rightarrow 0}\dfrac{\cos\;x-1}{x}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0}\dfrac{\dfrac{d}{\textit{dx}}( \cos\;x-1) }{\dfrac{d}{\textit{dx}}x}=\dfrac{1}{2} \lim\limits_{x\rightarrow 0}\dfrac{-\!\sin\;x}{1}=0\\[-1.8pc] &&\hspace{17pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }} } \end{eqnarray*} \]