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 :)
 A: $$ \lim_{x\to 0}{\frac{x-\arcsin{x}}{\sin^{3}{x}}}=\lim_{x\to 0}{\left(-\frac{\arcsin{x}}{\sin{x}}\right)^{3}\frac{\arcsin{x}-x}{\arcsin^{3}{x}}}=\left(-1\right)^{3}\times\frac{1}{6}=-\frac{1}{6} $$
Because $$ \lim_{x\to 0}{\frac{\arcsin{x}}{\sin{x}}}=\lim_{x\to 0}{\frac{\arcsin{x}}{x}\frac{x}{\sin{x}}}=1\times 1=1 $$
And doing the change of variable $ y=\arcsin{x} $, we have : $$ \lim_{x\to 0}{\frac{\arcsin{x}-x}{\arcsin^{3}{x}}}=\lim_{y\to 0}{\frac{y-\sin{y}}{y^{3}}}=\frac{1}{6} $$
If you Don't know how to prove the well-known limit $ \lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{3}}}=\frac{1}{6} $, then here is my proof without using L'Hopital :
Let $ x\in\left[-\pi,\pi\right]\setminus\left\lbrace 0\right\rbrace $, observe that : $$ \fbox{$\begin{array}{rcl}\displaystyle\frac{x-\sin{x}}{x^{3}}=\frac{1}{2}\int_{0}^{1}{\left(1-t\right)^{2}\cos{\left(tx\right)}\,\mathrm{d}t}\end{array}$} $$
Since $ \frac{1}{6}=\frac{1}{2}\int\limits_{0}^{1}{\left(1-t\right)^{2}\,\mathrm{d}t} $, we have the following : $$ \left|\frac{1}{6}-\frac{x-\sin{x}}{x^{3}}\right|=\frac{1}{2}\int_{0}^{1}{\left(1-t\right)^{2}\left(1-\cos{\left(tx\right)}\right)\mathrm{d}t} $$
Since $ \left(\forall y\in\left[-\pi,\pi\right]\right),\ \frac{y^{2}}{2}+\cos{y}-1=\frac{y^{3}}{2}\int\limits_{0}^{1}{\left(1-x\right)^{2}\sin{\left(xy\right)}\,\mathrm{d}x}\geq 0 $, we get in particular : $$ \left(\forall t\in\left[0,1\right]\right),\ 1-\cos{\left(tx\right)}\leq\frac{\left(tx\right)^{2}}{2} $$ And thus : \begin{aligned} \left|\frac{1}{6}-\frac{x-\sin{x}}{x^{3}}\right|&\leq\frac{x^{2}}{4}\int_{0}^{1}{t^{2}\left(1-t\right)^{2}\,\mathrm{d}t}\\ \iff \left|\frac{1}{6}-\frac{x-\sin{x}}{x^{3}}\right|&\leq\frac{x^{2}}{120}\end{aligned}
Meaning, we have : $ \lim\limits_{x\to 0}{\frac{x-\sin{x}}{x^{3}}}=\frac{1}{6}\cdot $
A: 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
$$
\lim_{x\to0}\frac{x-x-x^3/6+o(x^3)}{(x+o(x))^3}=-\frac{1}{6}
$$
A: L'Hospital's rule is not the alpha and omega of limits computation! It is very often  simpler to use an equivalent  of the numerator and the denominator:


*

*For the denominator, it is quite simple: $\sin x\sim_0 x$, so $\sin^3x \sim_0 x^3$.

*For the numerator, you have to know the Taylor expansion of  arcsine:
$$\arcsin x=x+\frac{x^3}6+o(x^3),\quad\text{so }\; x-\arcsin x=-\frac{x^3}6+o(x^3) $$
which means the numerator is equivalent near $0$ to $-\frac{x^3}6$, and the given fraction
$$\frac{x-\arcsin x}{\sin^{3}x}\sim_0\frac{-\frac{x^3}6}{x^3}=-\frac16.$$
A: If you cannot with L'Hôpital, you can , for example developpe in series so you get the quotient. But first change variable $\arcsin (x)=y$ so you want at neighborhood of zero
$$\frac{sin(y)-y}{\sin^3(\sin(y))}=\dfrac{-\dfrac {y^3}{6}+\dfrac{y^5}{120}+O(y^7)}{y^3-y^5+O(y^7)}$$ Then the limit is $-\dfrac16$
