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.

Without using L'Hospital's and Series Expansion,
Using this and Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$, 
$$\arcsin x-\arctan x=\arcsin x-\arcsin\frac x{\sqrt{1+x^2}}=\arcsin\left(x\cdot\frac1{\sqrt{1+x^2}}-\sqrt{1-x^2}\cdot\frac x{\sqrt{1+x^2}} \right)$$
$$=\arcsin\left(x\cdot\frac{(1-\sqrt{1-x^2})}{\sqrt{1+x^2}}\right)$$
$$=\arcsin\left(x\cdot\frac{1-(1-x^2)}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)=\arcsin\left(\frac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)$$
$$\implies\lim_{x\to 0}\frac{\arcsin(x)-\arctan(x)}{x^3}$$
$$=\lim_{x\to 0}\frac{\arcsin\left(\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)}{\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}}\cdot\frac1{\lim_{x\to 0}\sqrt{1+x^2}(1+\sqrt{1-x^2})}$$
Set $\displaystyle\arcsin\left(\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}\right)=\theta\implies\dfrac{x^3}{\sqrt{1+x^2}(1+\sqrt{1-x^2})}=\sin\theta$ in the first limit to find the limit to be $1$
The second limit is easy enough
