TITLE: Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$
QUESTION [8 upvotes]: Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):
$$ \lim_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}? $$

REPLY [4 votes]: Letting $y = x^3$, note that your limit is the same as
$$
  \lim_{y \rightarrow 0} \frac{\sin(y^{1 \over 3}) - y^{1 \over 3}\cos(y^{1 \over 3})}{y}
$$
If you let the mean value theorem be one of your standard rules, then you can say that ${\displaystyle \frac{\sin(y^{1 \over 3}) - y^{1 \over 3}\cos(y^{1 \over 3})}{y} = f'(z)}$ for some $z$ between $0$ and $y$. Calculating the derivative reveals that
$$
  f'(z) = {1 \over 3}{\sin(z^{1 \over 3}) \over z^{1 \over 3}}
$$
Observe that as $y$ goes to zero, so does $z$, and therefore so does $z^{1 \over 3}$. Using that ${\displaystyle \lim_{x \rightarrow 0} {\sin(x) \over x} = 1}$ one sees therefore that
$$
  \lim_{y \rightarrow 0} f'(z) = {1 \over 3}
$$
So ${1 \over 3}$ is the limit.