Question

$\underset{x\to 1^{-}}{lim}\frac{\sqrt{\pi }-\sqrt{2{\sin }^{-1}x}}{\sqrt{1-x}}$ is equal to :

• A. $\sqrt{\frac{2}{\pi }}$
• B. $\sqrt{\frac{\pi }{2}}$
• C. $\frac{1}{\sqrt{2\pi }}$
• D. $\sqrt{\pi }$

Answer 1

The correct option is A $\sqrt{\frac{2}{\pi }}$

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

$=\underset{x\to 1^{-}}{lim}\frac{\pi -2{\sin }^{-1}x}{\sqrt{1-x}(\sqrt{\pi }+\sqrt{2{\sin }^{-1}x})}$

$=\frac{1}{2\sqrt{\pi }}\underset{x\to 1^{-}}{lim}\frac{\pi -2{\sin }^{-1}x}{\sqrt{1-x}}\left[\frac{0}{0}\text{form}\right]$

Applying L'Hospital Rule, we get
$\frac{1}{2\sqrt{\pi }}\times \underset{x\to 1^{-}}{lim}\frac{\frac{-2}{\sqrt{1-x^2}}}{\frac{-1}{2\sqrt{1-x}}}$

$=\frac{1}{\sqrt{\pi }}\times \underset{x\to 1^{-}}{lim}\frac{2}{\sqrt{1+x}}$

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