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

Hint:

Let $\arcsin x= t\implies x=\sin t,\dfrac\pi4\le t\le\dfrac\pi2,\cos t=+\sqrt{1-x^2}$

$$\dfrac{\cos^{-1}(\sin2t)}{\sin t-\dfrac1{\sqrt2}}=\dfrac{\dfrac\pi2-\sin^{-1}(\sin2t)}{\sin t-\dfrac1{\sqrt2}}$$

Now using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,

$\sin^{-1}(\sin2t)=\pi-2t$ as $2t\ge\dfrac\pi2$

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