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?

$$\frac{1}{\arcsin{x}}-\frac{1}{\sin{x}}=\frac{1}{x+\frac{x^3}{6}+...}-\frac{1}{x-\frac{x^3}{6}+...}\rightarrow0.$$
Your way  also helps because $$\frac{\sin{x}-\arcsin{x}}{x^2}\rightarrow\frac{\cos{x}-\frac{1}{\sqrt{1-x^2}}}{2x}\rightarrow\frac{-\sin{x}-\frac{x}{\sqrt{(1-x^2)^3}}}{2}\rightarrow0.$$
