# $\lim_{x\to0}\frac{\arcsin x-\sin x}{x^3}$ 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 $$2.10$$

• You will only need to apply L'Hospital twice – Shubham Johri Feb 11 at 17:54
• @ShubhamJohri Doesn't the whole thing look ugly? – tatan Feb 11 at 17:56
• It's as easy with L'Hospital as any answer below – Shubham Johri Feb 11 at 18:01
• @ShubhamJohri Maybe. But the answer below is more elegant. – tatan Feb 11 at 18:07
• – lab bhattacharjee Feb 11 at 18:25

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 $$\frac{1}{3}$$.
You could substitute $$\arcsin x=\alpha$$ with $$\alpha\to0$$, you will only require the series for $$\sin y$$ then.
$$\lim_{\alpha\to0}\frac{\alpha-\sin\sin\alpha}{\sin^3\alpha}= \lim_{\alpha\to0}\frac{\alpha-\sin\alpha+\frac{\sin^3\alpha}{3!}...}{\sin^3\alpha}=\frac16+\lim_{y\to0}\frac{y-\sin y}{\sin^3y}=\frac16+\lim_{y\to0}\frac{y-y+\frac{y^3}{3!}...}{\sin^3y}=1/3$$