Find the limit: limit x tends to zero [arcsin(x)-arctan(x)]/(x^3) I'm having difficulty in finding the following limit.
$$\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
I tried manipulating the given limit in standard limit(s) but I got nowhere. I tried L'Hôpital's rule and then realized it would be very lengthy and daunting so I left it. The answer is 1/2.
Thanks for any help in advance.

It's worth mentioning that l'Hopital does work here. Your limit becomes
$$\lim_{x \rightarrow 0} {{1 \over \sqrt{1 - x^2}} - {1 \over 1 + x^2} \over 3x^2}$$
$$= \lim_{x \rightarrow 0}{(1 + x^2) - \sqrt{1 - x^2} \over 3x^2(1 + x^2)\sqrt{1 - x^2}}$$
Multiply numerator and denominator by $(1 + x^2) + \sqrt{1 - x^2}$ and you get
$$= \lim_{x \rightarrow 0}{(1 + x^2)^2 - (1 - x^2) \over 3x^2 ((1 + x^2) + \sqrt{1 - x^2})(1 + x^2)\sqrt{1 - x^2}}$$
$$= \lim_{x \rightarrow 0}{3  + x^2 \over 3((1 + x^2) + \sqrt{1 - x^2})(1 + x^2)\sqrt{1 - x^2}}$$
You can now just plug in $x = 0$ into the above and get ${1 \over 2}$.
