# 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.

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

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.

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

• @tharanga dharmapala Any further doubt? – Robert Z Sep 8 '17 at 13:02

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: $$\lim_{x\to \pi/2^+} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^+} \frac{-\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=-\sqrt{2\pi}.$$ $$\lim_{x\to \pi/2^-} \frac{|\sin{(\pi-2x)|}\cdot\sqrt{2\pi}}{(\pi-2x)}=\lim_{x\to \pi/2^-} \frac{\sin{(\pi-2x)}\cdot\sqrt{2\pi}}{(\pi-2x)}=\sqrt{2\pi}.$$

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 $t\to 0$ $$=\sqrt{2π} \lim_{t\to 0} \frac {\sin t}{t}$$

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