# Find the limit $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$

How can I find $\lim_{x\to0}\frac{\arcsin x -x}{x^2}$?

I've tried using the Lhopital rule and it got me here:

$\lim_{x\to0}\frac{\arcsin x -x}{x^2} = \lim_{x\to0}\frac{\frac{1}{\sqrt{1-x^2}}-1}{2x} = \lim_{x\to0}\frac{x}{2\sqrt{1-x^2}\cdot (1+\sqrt{1-x^2})}$

This doesn't make life much easier, unless I could say that $\frac{x}{2\sqrt{1-x^2}\cdot (1+\sqrt{1-x^2})}$ is continuous at $x=0$..

Is there a better way to approach this?