Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$

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 $$\lim_{x\to 0} \frac{\sin(x)-x\cos(x)}{x^3} =\lim_{x\to 0} \frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}=\frac{1}{3}$$

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.