Question

Evaluate $\underset{x\to \frac{\pi }{4}}{lim}\frac{\sqrt{1-\sqrt{\sin 2x}}}{\pi -4x}$

• A. $-\frac{1}{2}$
• B. $-\frac{1}{4}$
• C. $\frac{1}{4}$
• D. does not exists

Answer 1

The correct option is D does not exists

$\underset{x\to \pi /4}{lim}\frac{\sqrt{1-\sqrt{\sin 2x}}}{\pi -4x}$
$LHL=\underset{h\to 0}{lim}\frac{\sqrt{1-\sqrt{\sin \left[\frac{\pi }{2}-2h\right]}}}{\pi -\pi +4h}$
$=\underset{h\to 0}{lim}\frac{\sqrt{1-\sqrt{\cos 2h}}}{4h}\times \frac{\sqrt{1+\sqrt{\cos 2h}}}{\sqrt{1+\sqrt{\cos 2h}}}$
$=\underset{h\to 0}{lim}\frac{\sqrt{1-\sqrt{\cos 2h}}}{4h}\times \frac{1}{\sqrt{1+\sqrt{\cos 2h}}}$
$=\frac{\sqrt{2}|\sin h|}{4h\times \sqrt{2}}=\frac{1}{4}$
Similarly $RHL=-\frac{1}{4}$
Hence $LHL\ne RHL$ $\therefore$ limit does not exist.