<s>[INST] Find $\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$ $$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$$
How do I go about doing this? I can see no simple way of doing this. Application of l'Hopital's rule would be very laborious. A Taylor expansion seems feasible but is that the best way? It seems like it may be also very laborious.
 [/INST] I may use classic limits (which can easily be computed by hospital's rule
(twice)
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2}
\end{eqnarray*}
The idea is to write the original expression using those above,
\begin{equation*}
\frac{\sin x-x\cos x}{x^{3}}=\frac{\sin x-x+x-x\cos x}{x^{3}}=\frac{\sin x-x%
}{x^{3}}+\frac{1-\cos x}{x^{2}}
\end{equation*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x-x\cos x}{x^{3}}=\lim_{x\rightarrow 0}\frac{%
\sin x-x}{x^{3}}+\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=-\frac{1}{6}+%
\frac{1}{2}=\frac{1}{3}.
\end{equation*}
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