Limits problem $ \lim_{x\to \pi^+/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$ $$ \lim_{x\to \pi^+/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$$ 
I tried to solve this by L'hostipal's rule..but that doesn't give a solution..appreciate if you can give a clue..
$ \lim_{x\to \pi^+/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$
 A: \begin{align}
 \lim_{x\to \frac{\pi^+}{2}} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}&=  \lim_{x\to \frac{\pi^+}{2}} \frac{\sqrt{2\cos^2x}}{\sqrt{\pi}-\sqrt{2x}}\\
&=  \lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}\cos x}{\sqrt{\pi}-\sqrt{2x}}\\
&=  \lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}\cos x(\sqrt{\pi}+\sqrt{2x})}{\pi-2x}\\
&=  \lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}\sin (\frac{\pi}{2}-x)(\sqrt{\pi}+\sqrt{2x})}{2(\frac{\pi}{2}-x)}\\
&=  \lim_{x\to \frac{\pi^+}{2}} \frac{\sin (\frac{\pi}{2}-x)}{(\frac{\pi}{2}-x)}\lim_{x\to \frac{\pi^+}{2}} \frac{-\sqrt{2}(\sqrt{\pi}+\sqrt{2x})}{2}\\
&=  (1)\left[\frac{-\sqrt{2}(\sqrt{\pi}+\sqrt{\pi})}{2}\right]\\
&=-\sqrt{2\pi}
\end{align}
A: Let $t=x-\frac\pi2$,
$$\lim_{x\to\frac\pi2^+}\frac{\sqrt{1+\cos 2x}}{\sqrt\pi-\sqrt{2x}}$$
$$=\lim_{t\to0^+}\frac{\sqrt{1-\cos2t}(\sqrt\pi+\sqrt{2t+\pi})}{(\sqrt\pi+\sqrt{2t+\pi})(\sqrt\pi-\sqrt{2t+\pi})}$$
$$=\lim_{t\to0^+}\frac{\sqrt{1-(1-2\sin^2 t)}(\sqrt\pi+\sqrt{2t+\pi})}{-2t}$$
$$=\lim_{t\to0^+}\frac{\sqrt{2}\sin t(\sqrt\pi+\sqrt{2t+\pi})}{-2t}$$
$$=-\sqrt{2\pi}$$
