<s>[INST] 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?
 [/INST] 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 $$\lim_{x\to\frac1{\sqrt2}^-}\frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}=-2\lim_{x\to\frac1{\sqrt2}^-}\frac{\arcsin x-\dfrac\pi4}{x-\frac{1}{\sqrt{2}}}=?$$
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