Why series expansion in evaluating $\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}=\sqrt2$ is not working here? Prove that $\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}=\sqrt2$ 

I can evaluate this limit using L Hospital rule but i wonder why Series expansion method is failing here and L Hospital rule is working here. 
$\lim_{x\to 0^+}\frac{\arccos(1-x^2)}{x}$
As $\arccos(x)=\frac{\pi}{2}-x-\frac{x^3}{6}-\frac{3x^5}{40}-....$
$\arccos(1-x^2)=\frac{\pi}{2}-(1-x^2)-\frac{(1-x^2)^3}{6}-\frac{3(1-x^2)^5}{40}-....$
$\frac{\frac{\pi}{2}-(1-x^2)-\frac{(1-x^2)^3}{6}-\frac{3(1-x^2)^5}{40}-....}{x}$
And it is not simplifying.I cannot take $x$ common from the numerator to cancel out from the denominator.
Have i made some mistake or series expansion will not work here.
Are there some cases where series expansion fails and only L Hospital works.Please help me.Thanks.
 A: As $x\to0,1-x^2>0$
$\arccos(1-x^2)=\arcsin\sqrt{1-(1-x^2)^2}=\arcsin\sqrt{2x^2-x^4}$
$$\lim_{x\to0^+}\dfrac{\arcsin\sqrt{2x^2-x^4}}x=\lim_{x\to0^+}\dfrac{\arcsin\sqrt{2x^2-x^4}}{\sqrt{2x^2-x^4}}\cdot\lim_{x\to0^+}\dfrac{x\sqrt{2-x^2}}x$$
Set $\arcsin\sqrt{2x^2-x^4}=y$ in the first
Can you take it home from here?
A: HINT:
Let $\arccos(1-x^2)=2y\implies1-x^2=\cos2y\iff x^2=1-\cos2y=2\sin^2y$
As $0\le2y\le\pi, \sin y\ge0$ 
and $x>0, x=\sqrt2\sin y$
Can you take it home from here?
A: Actually, we can expand the arccosine directly.  

Relatedly: 
Note that in This Answer, I showed a way to expand $\arcsin(x)$ around $x=1$. 

Let $f(x)=\arccos(1-x^2)$.  Then, for $x\ne 0$
$$\begin{align}
f'(x)&=\frac{2x}{\sqrt{1-(1-x^2)^2}}\\\\
&=\frac{2x}{\sqrt{2x^2-x^4}}\\\\
&=\frac{2x}{|x|}\left(2-x^2\right)^{-1/2}\\\\
&=2\text{sgn}(x)\left(2-x^2\right)^{-1/2}
\end{align}$$
Note that $\displaystyle \lim_{x\to 0^{\pm}}f'(x)=\pm \sqrt{2}$.  We find, therefore, that 
$$f(x)=\sqrt{2}|x|+O\left(|x|^3\right)$$
and the coveted limit is given by 
$$\lim_{x\to 0^+}\,\frac{\arccos(x)}x=\lim_{x\to 0^+}\left(\frac{\sqrt{2}\,\,|x|+O\left(|x|^3\right)}x\right)=\sqrt 2$$
as was to be shown!
