Question

Solve it:
$\underset{x\to \frac{\pi }{2}}{lim}\frac{\sqrt{2-\sin x-}1}{\left(\frac{\pi }{2}-x\right)}$

Answer 1

Solution -

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\sqrt{2-sinx-1}}{(\frac{\pi }{2}-x)}$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\sqrt{1-sinx}}{\frac{\pi }{2}-x}$

Apply L-Hospital rule

$\underset{x\to \frac{\pi }{2}}{lim}\frac{-cosx}{-2\sqrt{1-sinx}}$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{cos\sqrt{1+sinx}}{2\sqrt{1-sinx}\sqrt{1+sinx}}$

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\sqrt{1+sinx}}{2}$

$=\frac{\sqrt{2}}{2}$

$=\frac{1}{\sqrt{2}}$