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?

That $\sqrt{1-x^2}$ shouts "Trigonometric functions!!!". So:
Put $x=\sin\theta$, which means that $\sqrt{1-x^2}=\cos\theta$. Then $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\arccos \left(\sin 2\theta \right)}{\sin \theta-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\frac{\pi}{2}-2\theta}{\sin \theta-\frac{1}{\sqrt{2}}}$$
You can use l'Hôpital from here. Or...
Put $\phi=\theta-\frac{\pi}{4}$. The limit becomes $$\lim_{\phi \to 0} \frac{-2\phi}{\sin \phi \cos\frac{\pi}{4} + \cos \phi \sin\frac{\pi}{4}-\frac{1}{\sqrt{2}}} \equiv \lim_{\phi \to 0} \frac{-2\phi}{\frac{1}{\sqrt{2}}(\sin \phi + \cos \phi -1)}$$
…which you can l'Hôpital again, or note that for small $\phi$, $\sin\phi$ is practically $\phi$ and $\cos\phi$ is practically 1, so that the limit becomes $$\frac{-2}{\frac{1}{\sqrt{2}}}$$
…which I leave to you to work out. This is essentially reproducing what l'Hôpital does, but by doing it yourself you can actually see what you are doing.
