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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.

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$\cos{2x}=2\cos^2{x}-1$

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.

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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 $$\lim_{x\to (\pi/2)^{\pm}} \frac{\sqrt{1+\cos(2x)}}{\sqrt{\pi}-\sqrt{2x}}=\mp\sqrt{2\pi}.$$

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  • $\begingroup$ @tharanga dharmapala Any further doubt? $\endgroup$ – Robert Z Sep 8 '17 at 13:02
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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: $$\lim_{x\to \pi/2^+} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^+} \frac{-\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=-\sqrt{2\pi}.$$ $$\lim_{x\to \pi/2^-} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^-} \frac{\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=\sqrt{2\pi}.$$

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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}$$

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

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