trigonometry - Limit and L'hopital's Rule

How do you evaluate the following limit?
$$\lim_{x \to 0} \dfrac{x\sin^{-1}x}{x-\sin{x}}$$
When I is L'Hopital's rule twice, I get:
$$\lim_{x \to 0} \dfrac{(x^2+2)\csc x}{(1-x^2)^{3/2}}$$
Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.

So, how do I find this limit?

Answer

$$\lim_{x\to0}\dfrac{x\sin^{-1}x}{x-\sin x}=\lim_{x\to0}\dfrac{\sin^{-1}x}x\cdot\lim_{x\to0}\dfrac1{\dfrac{x-\sin x}{x^3}}\cdot\lim_{x\to0}\dfrac1x$$

The first & the last limits are elementary and

for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion