Question

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

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

Answer 1

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

$L=\underset{x\to -1}{lim}\frac{\sqrt{\pi }-\sqrt{{\cos }^{-1}x}}{\sqrt{x+1}}$
Clearly form of the limit is, $\frac{0}{0}$, thus applying L-Hospital's rule,
$L=\underset{x\to -1}{lim}\frac{\frac{1}{2\sqrt{{\cos }^{-1}x}}.\frac{1}{\sqrt{1-x^2}}}{\frac{1}{2\sqrt{x+1}}}=\underset{x\to -1}{lim}\frac{1}{\sqrt{{\cos }^{-1}x}}.\frac{1}{\sqrt{1-x}}=\frac{1}{\sqrt{2\pi }}$