Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$ 
$$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$

I was doing some questions on limits, I saw one in which there is $\arccos x$. I am stuck there, not able to proceed. 
Can you give me some hint?

Set $t=\arccos(2x\sqrt{1-x^2})$, so $2x\sqrt{1-x^2}=\cos t$ and, by definition, $0\le t\le\pi$. Then
$$
\sin t=\sqrt{1-\cos^2t}=\sqrt{1-4x^2+4x^4}=|2x^2-1|
$$
and so $t=\arcsin|2x^2-1|$. Thus your limit (from the right) can be written
$$
\lim_{x\to(1/\sqrt2)^+}\sqrt{2}\frac{\arcsin|2x^2-1|}{\sqrt{2}x-1}=
\lim_{x\to(1/\sqrt2)^+}\sqrt{2}
\frac{(\sqrt{2}x+1)\arcsin(2x^2-1)}{2x^2-1}
$$
which you should be able to compute.
Note then that, for $x<1/\sqrt{2}$, $|2x^2-1|=-(2x^2-1)$ and you can go similarly for the limit as $x\to(1/\sqrt{2})^-$
