Limit of a trig function. (Without using L'Hopital) I'm having trouble figuring out what to do here, I'm supposed to find this limit:
$$\lim_{x\rightarrow0} \frac{x\cos(x)-\sin(x)}{x^3}$$
But I don't know where to start, any hint would be appreciated, thanks!

One can use L'Hospital's rule or, as shown here, the series expansion of $\sin(x)$ and $\cos(x)$. 
Using 
\begin{align}
\cos(x) &= 1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} - \cdots \\
\sin(x) &= x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots
\end{align}
then
$$x \cos(x) - \sin(x) = - \frac{2 \, x^{3}}{3!} + \frac{4 \, x^{5}}{5!} - \cdots$$
and
$$\lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{x^{3}} = \lim_{x \to 0} \left( - \frac{2}{3!} + \frac{4 \, x^{2}}{5!} - \cdots \right) = - \frac{1}{3}.$$
