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?
 A: 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 $x\to(1/\sqrt{2})^-$
A: First of all, $$\arccos2x\sqrt{1-x^2}=\dfrac\pi2-\arcsin2x\sqrt{1-x^2}$$
Now using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $, $$2\arcsin x=\begin{cases}\arcsin2x\sqrt{1-x^2} &\mbox{if } |x|\le\dfrac1{\sqrt2} \\ \pi-\arcsin2x\sqrt{1-x^2}  & \mbox{if } x>\dfrac1{\sqrt2}\\-\pi-\arcsin2x\sqrt{1-x^2}  & \mbox{if } x<-\dfrac1{\sqrt2}
 \end{cases}$$
$$\implies\lim_{x\to\frac1{\sqrt2}^+}\frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}=2\lim_{x\to\frac1{\sqrt2}^+}\frac{\arcsin x-\dfrac\pi4}{x-\frac{1}{\sqrt{2}}}=\dfrac{d(\arcsin x)}{dx}_{(\text{ at } x=\pi/4)}$$
and $$\lim_{x\to\frac1{\sqrt2}^-}\frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}=-2\lim_{x\to\frac1{\sqrt2}^-}\frac{\arcsin x-\dfrac\pi4}{x-\frac{1}{\sqrt{2}}}=?$$
A: 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.
