Question

$\underset{x\to -1^{+}}{lim}\frac{\sqrt{\pi }-\sqrt{{\cos }^{-1}x}}{\sqrt{x+1}}$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{2\pi }}$
• C. $\frac{1}{\sqrt{\pi }}$
• D. none of these

Answer 1

The correct option is B $\frac{1}{\sqrt{2\pi }}$

Substitute $y={\cos }^{-1}x,$ so that $x=\cos y$ and as $x\to -1,y\to \pi$

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

$=\underset{y\to {\pi }^{+}}{lim}\frac{\sqrt{\pi }-\sqrt{y}}{\sqrt{2}\cos \left(y/2\right)}$

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

$=\underset{y\to {\pi }^{+}}{lim}\frac{2}{\sqrt{2}}.\frac{\pi /2-y/2}{\sin (\pi /2-y/2)}.\frac{1}{\sqrt{\pi }+\sqrt{y}}$
$=\frac{1}{\sqrt{2\pi }}[\because \underset{u\to 0}{lim}=1,takingu=\frac{\pi }{2}-\frac{y}{2}]$