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.
 A: Taylor expanding we find that:
\begin{equation}
\begin{aligned}
\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}&=\lim_{x\to 0}\frac{x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)-x\left[1-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right]}{x^3}\\
&=\lim_{x\to 0}\frac{x^3\left(\frac{1}{2}-\frac{1}{6}\right)+\mathcal{O}\left(x^4\right)}{x^3}\\
&=\frac{1}{3}
\end{aligned}
\end{equation}
A: Use l'Hopital, not laborious at all..
$$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3} = \lim_{x\to 0}\frac{\cos x-(\cos x - xsinx)}{3x^2} = \lim_{x\to 0}\frac{sinx}{3x} = \frac{1}{3} $$
A: 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*}
A: Taylor series
$$ \sin x = x - \frac 16 x^3 + O(x^5)$$
$$ x \cos x = x - \frac 12 x^3 + O(x^5)$$
so 
$$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3} = \lim_{x\to 0}
\left(\frac 13 + O(x^2)\right) = \frac 13$$
