Question

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

Answer 1

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

Put $co{s}^{-1}x=t\Rightarrow x=cost$

As $x\to -1,\therefore t\to \pi$

$=\underset{t\to \pi }{lim}\frac{\sqrt{\pi }-\sqrt{t}}{\sqrt{1+cost}}$

$=\underset{t\to \pi }{lim}\frac{(\pi -t)}{\sqrt{1-co{s}^{2}t}}\frac{\sqrt{1-cost}}{(\sqrt{\pi }+\sqrt{t})}$

$=\underset{t\to \pi }{lim}\frac{\pi -t}{sint}\frac{\sqrt{1+1}}{2\sqrt{\pi }}$

$=\underset{t\to \pi }{lim}\frac{\pi -t}{sin(\pi -t)}\times \frac{\sqrt{2}}{2\sqrt{\pi }}$

$=\frac{1}{\sqrt{2\pi }}$