Question

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

• A. $0$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{2\pi }}$
• D. $\sqrt{\frac{\pi }{2}}$

Answer 1

The correct option is A $0$

Given,

${lim}_{x\to 1+}\left(\frac{\sqrt{\pi }\sqrt{{\cos }^{-1}\left(x\right)}}{\sqrt{x+1}}\right)$

put $x=1$

$=\frac{\sqrt{\pi }\sqrt{{\cos }^{-1}\left(1\right)}}{\sqrt{1+1}}$

$=\frac{\sqrt{\pi }\sqrt{0}}{\sqrt{2}}$

$=0$