<s>[INST] Limit calculation (L'Hospital?) I need to calculate the following limit:
$$
\lim_{x\to0}{\lim}\frac{x-\arcsin\left(x\right)}{\sin^{3}(x)}
$$
but I have no idea how to do it. I tried to use L'Hospital (as it meets the conditions), but it looks like it's only get worse:
\begin{align}\lim_{x\to0}\frac{f'\left(x\right)}{g'\left(x\right)}&=\lim_{x\to 0}\frac{1-\frac{1}{\sqrt{1-x^{2}}}}{3\sin^{2}\left(x\right)\cos\left(x\right)}=\lim_{x\to 0}\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}\cdot\left(3\sin^{2}\left(x\right)\cos\left(x\right)\right)}\\
&=\lim{x\to 0}\biggl(1-\frac{1}{\sqrt{1-x^{2}}}\biggr)\biggl(\frac{1}{3\sin^{2}\left(x\right)\cos\left(x\right)}\biggr)
\end{align}
Someone told me to apply L'Hospital again (and maybe it could work), but it gets really complicated.
Maybe there is another (more simple) way? 
Thanks in advance :)
 [/INST] You can first rewrite the limit as
$$
\lim_{x\to0}\frac{x-\arcsin x}{x^3}\frac{x^3}{\sin^3x}
$$
Since the last fraction has limit $1$, you can concentrate on
$$
\lim_{x\to0}\frac{x-\arcsin x}{x^3}=
\lim_{x\to0}\frac{1-\dfrac{1}{\sqrt{1-x^2}}}{3x^2}=
\lim_{x\to0}\frac{\sqrt{1-x^2}-1}{x^2}\frac{1}{3\sqrt{1-x^2}}
$$
The second fraction has limit $1/3$, so you can concentrate on
$$
\lim_{x\to0}\frac{\sqrt{1-x^2}-1}{x^2}=\lim_{x\to0}\frac{-\dfrac{x}{\sqrt{1-x^2}}}{2x}=-\frac{1}{2}
$$
by an easy verification.
Hence the global limit is 
$$
1\cdot\frac{1}{3}\cdot\left(-\frac{1}{2}\right)=-\frac{1}{6}
$$

If you don't remember the Taylor expansion of the arcsine, it is not difficult to find it up to degree $3$, which is all it's needed here.
Indeed, the derivative is $(1-x^2)^{-1/2}$, so the Taylor expansion up to degree $2$ is
$$
1-\frac{1}{2}(-x^2)+o(x^2)=1+\dfrac{1}{2}x^2+o(x^2)
$$
Hence the Taylor expansion of the arcsine is
$$
c+x+\frac{1}{6}x^3+o(x^3)
$$
and $c=0$ because $\arcsin 0=0$. So your limit is
$$
\lim_{x\to0}\frac{x-x-x^3/6+o(x^3)}{(x+o(x))^3}=-\frac{1}{6}
$$
</s>