$ \lim_{x\to \frac{1}{{\sqrt 2}^+}} \frac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\frac{1}{\sqrt{2}}}$ 
$\displaystyle \lim_{x\to {1\over \sqrt{2}^+}} \dfrac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\dfrac{1}{\sqrt{2}}}$

I have tried substituting $x$ for $\sin \theta$, doing the calculations and ended up with -$2√2$. But the solution provided was $2√2$. Then I tried this question again, but this time used $\cos \theta$ instead of $\sin \theta$ and the answer did match. I don't understand why $x$ as $\sin \theta$ doesn't give the correct result. I have checked all my steps but couldn't find any flaw with $\sin \theta$ as substitution. Can anyone tell me whether $\sin \theta $ a wrong substitution for this question or not?

One may use l'Hopital rule, then
$$\lim_{x\to {1\over \sqrt{2}^+}} \dfrac{\cos ^{-1} \left( 2x\sqrt{1-x^2}\right)}{x-\dfrac{1}{\sqrt{2}}}  = \lim_{x\to {1\over \sqrt{2}^+}}\dfrac{-2(1-2x^2)}{\sqrt{1-x^2}\sqrt{1-4x^2-4x^4}} = \lim_{x\to {1\over \sqrt{2}^+}}\dfrac{-2(1-2x^2)}{\sqrt{1-x^2}(2x^2-1)}=2\sqrt{2}$$
