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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.

diya
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    You are using a series expansion around $0$. When plugging in $1-x^2$, the (infinite) series is still exact,but you cannot truncate it and keep only the first few terms: this would be only valid if $1-x^2\simeq 0$, which is not the case when $x\to0$. – Clement C. Dec 19 '15 at 16:07
  • Should i expand it around $1$,as $1-x^2\to 1$ as $x\to 0$.Will then it work and will that be OK.@ClementC. – diya Dec 19 '15 at 16:14

3 Answers3

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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?

lab bhattacharjee
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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?

lab bhattacharjee
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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!

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Mark Viola
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