Question

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

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

Answer 1

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

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

Let, $x=1+h$ then,

$=\underset{h\to 0}{lim}\frac{\sqrt{\pi }-\sqrt{{\cos }^{-1}(1+h})}{\sqrt{2+h}}$

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

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

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