Evaluating $\lim\limits_{x \to 0} \frac {x - \tan^{-1}x} {x\sin x}$

Question

I am curretly working on the following question:

$$\lim\limits_{x \to 0} \frac {x - \tan^{-1}x} {x\sin x}$$

I was thinking of splitting up the limit such as

$$\lim\limits_{x \to 0} \frac{x}{x \sin x} - \lim\limits_{x \to 0} \frac {\tan^{-1}x}{x \sin x}$$

For the LHS, I was thinking of doing L'Hopital's rule and would get

$$\lim\limits_{x \to 0} \frac{1}{x \cos x+\sin x}$$

I feel as if I'm just going in circles and would like guidance towards heading in the right direction.

Answer 1

$$\lim_{x \to 0}\dfrac{x-\arctan x}{x^2}\cdot\lim_{x \to 0}\dfrac x{\sin x}$$

Now apply L’Hôpital for the first limit.

Answer 2

Use Taylor expansion: $$\lim\limits_{x \to 0} \frac {x - \tan^{-1}x} {x\sin x}=\lim\limits_{x \to 0} \frac {x - \left(x-\frac{x^3}{3}+\frac{x^5}{5}+O(x^7)\right)} {x\sin x}=\\ \lim\limits_{x \to 0} \frac {x\left(\frac{x}{3}-\frac{x^3}{5}+O(x^5)\right)} {\sin x}=0.$$