<s>[INST] 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.
 [/INST] Dividing numerator and denominator by $\sqrt{2}$
$$\lim_{x\to \frac{π}{2}} \frac {\sqrt{\frac{1+\cos 2x}{2}}}{\sqrt {\frac{π}{2}} -\sqrt { x}}$$
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Multiplying and dividing by $(\sqrt {\frac{π}{2} } + \sqrt {x})$
$$=\lim_{x\to \frac {π}{2}}(\sqrt {\frac{π}{2}} + \sqrt x)(\frac {\sqrt {\frac {1+ \cos 2x}{2}}}{\frac {π}{2} -x})$$
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As $\sqrt {\frac {1+\cos 2x}{2}} = \cos x$
$$=(\lim_{x\to \frac{π}{2}} \sqrt {\frac{π}{2}} +\sqrt{x})(\lim _{x\to \frac{π}{2}}\frac {\cos x}{\frac {π}{2} - x})$$
$$= \sqrt {2π} \lim_{x\to \frac{π}{2}} \frac {\sin (\frac {π}{2} - x)}{\frac {π}{2} - x}$$
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Let $t = \frac {π}{2} -x$
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As $x\to \frac{π}{2}$ then $t\to 0$
$$ =\sqrt{2π} \lim_{t\to 0} \frac {\sin t}{t}$$

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

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