Limit.....

Yes u all are Right............

\lim_{x \to \frac{\pi}{4}}\frac{\sqrt{1-\sqrt{sin2x}}}{(\pi-4x)}=\lim_{x \to \frac{\pi}{4}}\frac{\sqrt{1-\sqrt{sin2x}}}{4.(\frac{\pi}{4}-x)}$\\\\ Let $\frac{\pi}{4}-x=t$, Then as $x\rightarrow \frac{\pi}{4}$, then $t\rightarrow 0$\\\\ So $\lim_{t \to \0}\frac{\sqrt{1-\sqrt{cos2t}}}{4.t}=\lim_{t \to \0}\frac{\sqrt{1-\sqrt{cos2t}}}{4.t}\times\frac{\sqrt{1+\sqrt{cos2t}}}{\sqrt{1+\sqrt{cos2t}}}$\\\\ $=\lim_{t \to \0}\frac{\sqrt{1-cos2t}}{4t.\sqrt{1+\sqrt{cos2t}}}=\lim_{t \to \0}\sqrt{2}.\frac{|sint|}{4t.\sqrt{1+\sqrt{cos2t}}}$\\\\ $\lim_{t \to \0}\frac{1}{2.\sqrt{2}.\sqrt{1+\sqrt{cos2t}}}\times \underbrace{\lim_{t \to \0}\frac{|sint|}{t}}$\\\\ Here the Limit does not exist bcz L.H.L for $\lim_{t \to \0}\frac{|sint|}{t}=-1$\\\\ and R.H.L for $\lim_{t \to \0}\frac{|sint|}{t}=+1$.\\\\ So \boxed{\boxed{$\lim_{x \to \frac{\pi}{4}}\frac{\sqrt{1-\sqrt{sin2x}}}{(\pi-4x)}$= Does not exist.}}