How can I solve this limits: $\mathop {\lim }\limits_{x \to 0 } {{(\tan x - \sin x)\arctan x} \over {{1 \over 2}{x^4}}}$

Question

I know when $x \to 0$ , we have $sinx \sim arctanx \sim x $, but I have no idea when it comes to this form: $tanx -sinx$ , I'll appreciate it as long as a little hint.

Answer 1

Hint: $(x\to0)$

$$\eqalign{ & \tan x\sim\sin x\sim\arctan x\sim\arcsin x\sim x \cr & 1 - \cos x\sim{1 \over 2}{x^2} \cr & \tan x - \sin x = \tan x(1 - \cos x)\sim{1 \over 2}{x^3} \cr} $$

So the answer to the question is: $$\eqalign{ & \mathop {\lim }\limits_{x \to 0} {{(\tan x - \sin x)\arctan x} \over {{1 \over 2}{x^4}}} \cr & = \mathop {\lim }\limits_{x \to 0} {{{1 \over 2}{x^3} \cdot x} \over {{1 \over 2}{x^4}}} \cr & = 1 \cr} $$