Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$ Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):
$$ \lim_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}? $$

Use that 
$$\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots$$
and that 
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots$$
Hence
$$\frac{\sin(x)-x\cos(x)}{x^3}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}- x+\frac{x^3}{2}- \frac
{x^5}{4!} \pm \dots}{x^3}=\frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}\pm \dots$$
Hence the Limit of
$$\lim_{x\to 0} \frac{\sin(x)-x\cos(x)}{x^3} =\lim_{x\to 0} \frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}=\frac{1}{3}$$
