$\underset{x→\frac{1}{\sqrt{2}}}{\lim}\frac{\cos^{-1}(2x\sqrt{1-x^2})}{x-\frac{1}{\sqrt{2}}}$ is equal to

does not exist

Let x = cosθ i.e., as $x→\frac{1}{\sqrt{2}}$; $θ→\frac{π}{4}$

$RHL=\underset{θ→\frac{π}{4}^+}{\lim}\frac{\cos^{-1}(\sin 2θ)}{\cos θ-\frac{1}{\sqrt{2}}}=\underset{θ→\frac{π}{4}^+}{\lim}\frac{\cos^{-1}(\cos(\frac{π}{2}-2θ))}{\cos θ-\frac{1}{\sqrt{2}}}=\underset{θ→\frac{π}{4}^+}{\lim}\frac{2θ-\frac{π}{2}}{\cos θ-\frac{1}{\sqrt{2}}}=\underset{θ→\frac{π}{4}}{\lim}\frac{2}{-\sin θ}$

$=-2\sqrt{2}$

$RHL=\underset{θ→\frac{π}{4}^-}{\lim}\frac{\cos^{-1}(\sin 2θ)}{\cos θ-\frac{1}{\sqrt{2}}}=\underset{θ→\frac{π}{4}^-}{\lim}\frac{\cos^{-1}(\cos(\frac{π}{2}-2θ))}{\cos θ-\frac{1}{\sqrt{2}}}=\underset{θ→\frac{π}{4}^+}{\lim}\frac{\frac{π}{2}-2θ}{\cos θ-\frac{1}{\sqrt{2}}}$

$=\underset{θ→\frac{π}{4}^-}{\lim}\frac{-2}{-\sin θ}=2\sqrt{2}$