TITLE: How to calculate this limit as $x\rightarrow 0$?
QUESTION [5 upvotes]: Not sure how to evaluate this one any hints?$$\lim_{x\rightarrow 0} \frac{\tan^3(x)}{x}$$
I think I have an answer, would it be $0$. Since we can write is as $$\frac{\sin(x)}{x} \frac{\sin^2(x)}{\cos^3(x)}$$ and then use algebra of limit and established result for left hand limit?

REPLY [3 votes]: Your answer is correct. All it needs is to justify it, with something like

Because we know that if $\lim a_n$ and $\lim b_n$ exist, then $\lim a_nb_n=\lim a_n\lim b_n$, we can conclude that $$\lim \frac{\sin x}{x}\frac{\sin^2 x}{\cos ^3 x} = \lim \frac{\sin x}{x}\lim \frac{\sin ^2 x}{\cos^3 x} = 1\cdot 0 = 0.$$