limits - $lim_{xto0}frac{arcsin x-sin x}{x^3}$ without series or L'Hospital

$$\lim_{x\to0}\frac{\arcsin x-\sin x}{x^3}$$ without using series or L'Hospital

Is there any ohter simpler method? Expansion of $\arcsin$ is not trivial like tha of sine and L'Hospital is too cumbersome here.

Source-Question $2.10$

Answer

This site has repeatedly shown, without the methods forbidden in this question, that $\lim_{x\to 0}\frac{x-\sin x}{x^3}=\frac{1}{6}$. Hence $$\lim_{x\to 0}\frac{\arcsin x-x}{x^3}=\lim_{y\to 0}\frac{y-\sin y}{\sin^3 y}=\lim_{y\to 0}\frac{y-\sin y}{y^3}\left(\frac{y}{\sin y}\right)^3=\frac{1^3}{6}=\frac{1}{6}.$$Summing, your limit is $\frac{1}{3}$.