$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\
&= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}
But the answer is $\dfrac{1}{2}$ by L'Hopital's Rule.

Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.
We have 
$$\lim_{x \to 0} \tan x= \frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . .$$
$$\lim_{x \to 0} \sin x= \frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .$$
Therefore expression turns to,
$$\lim_{x \to 0} \frac{\frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . . -  (\frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .)}{x^{3}}$$
Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $\frac{1}{3}$ and $\frac{1}{6}$ which is $\color{red} {\frac{1}{2}}$
