Evaluate $ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$ 
Evaluate
  $$ \lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}$$ 

I tried to solve this by L'Hospital's rule..but that doesn't give a solution..appreciate if you can give a clue.
 A: The given limit does not exists. However we are able to evaluate the right and the left limit (without using L'Hospital's rule).
Note that 
$$\sqrt{1+\cos(2x)}=\sqrt{1+\cos^2(x)-\sin^2(x)}=\sqrt{2}|\cos(x)|=\sqrt{2}|\sin(\pi/2-x)|.$$
Hence as $x\to \pi/2$,
$$\frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}
=\frac{\sqrt{2}|\sin(\pi/2-x)|}{\sqrt{2}\frac{(\pi/2-x)}{\sqrt{\pi/2}+\sqrt{x}}}=\frac{|\sin(\pi/2-x)|}{\pi/2-x}\cdot \left(\sqrt{\pi/2}+\sqrt{x}\right).$$
Recalling that $\sin(t)/t$ goes to $1$ as $t\to 0$, we may conclude that
$$\lim_{x\to (\pi/2)^{\pm}} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}=\mp\sqrt{2\pi}.$$
A: $\cos{2x}=2\cos^2{x}-1$
Thus the nominator becomes:
$$\sqrt{1+\cos{2x}}=\sqrt{1+2\cos^2{x}-1}=\sqrt{2}|\cos{x}|$$
As $x \to \frac{\pi}{2}^-$ we have $|\cos{x}|=\cos{x}$
As $x \to \frac{\pi}{2}^+$ we have $|\cos{x}|=-\cos{x}$
Now you can use the L'Hospital rule to evaluate both limits.
A: Alternatively:
$$\lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}=\lim_{x\to \pi/2} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}\cdot \frac{\sqrt{1-\cos(2x)}}{\sqrt{1-\cos(2x)}}\cdot \frac{\sqrt{\pi}+\sqrt{2x}}{\sqrt{\pi}+\sqrt{2x}}=$$
$$\lim_{x\to \pi/2} \frac{|\sin{(2x)|}\cdot(\sqrt{\pi}+\sqrt{2x})}{(\pi-2x)\cdot\sqrt{1-\cos{(2x)}}}=\lim_{x\to \pi/2} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}$$
Note:
$$\lim_{x\to \pi/2^+} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^+} \frac{-\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=-\sqrt{2\pi}.$$
$$\lim_{x\to \pi/2^-} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^-} \frac{\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=\sqrt{2\pi}.$$
A: Dividing numerator and denominator by $\sqrt{2}$
$$\lim_{x\to \frac{π}{2}} \frac {\sqrt{\frac{1+\cos 2x}{2}}}{\sqrt {\frac{π}{2}} -\sqrt { x}}$$
$$$$
Multiplying and dividing by $(\sqrt {\frac{π}{2} } + \sqrt {x})$
$$=\lim_{x\to \frac {π}{2}}(\sqrt {\frac{π}{2}} + \sqrt x)(\frac {\sqrt {\frac {1+ \cos 2x}{2}}}{\frac {π}{2} -x})$$
$$$$
As $\sqrt {\frac {1+\cos 2x}{2}} = \cos x$
$$=(\lim_{x\to \frac{π}{2}} \sqrt {\frac{π}{2}} +\sqrt{x})(\lim _{x\to \frac{π}{2}}\frac {\cos x}{\frac {π}{2} - x})$$
$$= \sqrt {2π} \lim_{x\to \frac{π}{2}} \frac {\sin (\frac {π}{2} - x)}{\frac {π}{2} - x}$$
$$$$
Let $t = \frac {π}{2} -x$
$$$$
As $x\to \frac{π}{2}$ then $t\to 0$
$$ =\sqrt{2π} \lim_{t\to 0} \frac {\sin t}{t}$$

$$\lim_{x\to \frac{π}{2}} \frac {\sqrt {1+\cos 2x}}{\sqrt{π}-\sqrt 2x}=\sqrt {2π}$$

