$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
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.

Your problem arises from the fact that you used $\color{red}{\lim_\limits{x \to 0} \frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $\color{red}{\infty-\infty}$...
Only split an initial limit into a product if the individual limits are defined.
