<s>[INST] 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.
 [/INST] 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?
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