Find the limit $\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)$ 
Find the limit $$\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin
 x}\right)$$

\begin{align}
\lim_{x\to 0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)&=\lim_{x\to 0}\left(\frac{\sin x-\arcsin x}{\sin x\times \arcsin x}\right)\\
&=\lim_{x\to 0}\frac{x}{\arcsin x}
\cdot\frac{x}{\sin x}\cdot \frac{\sin x-\arcsin x}{x^2}
\end{align}
Here's where I'm stuck. I know from L'Hopital's Rule that $\frac{x}{\sin x}\to 1$, but what about $\frac{x}{\arcsin x}$? Or is there any other way than using L'Hopital to solve this question?

$$L=\lim_{x\to 0} \left( \frac{1}{\sin^{-1} x}-\frac{1}{\sin x}\right)$$
Note that if $|x|$ is very small then $\sin^{-1} x=x+x^3/6+..., \sin x=x-x^3/6+... (1+x)^{k}=1+kx+...$
Then $$L=\lim_{x \to 0} \left( \frac{1}{x+x^3/6}-\frac{1}{x-x^3/6} \right)$$
$$\implies L=\lim_{x \to 0}\frac{1}{x}[(1+x^2/6)^{-1}-(1-x^2/6)^{-1}]=\lim_{x \to 0}\frac{1}{x}[(1-x^2/6)-(1+x^2/6)]=0$$
