Mathematics - Limits Question with Solution | TestHub

Given limit can be written as

$L=\underset{x\to 1^{-}}{\lim }\frac{\sqrt{\pi }-\sqrt{2{\sin }^{-1}x}}{\sqrt{1-x}}\times \frac{\sqrt{\pi }+\sqrt{2{\sin }^{-1}x}}{\sqrt{\pi }+\sqrt{2{\sin }^{-1}x}}$

$\Rightarrow L=\underset{x\to 1^{-}}{\lim }\frac{\pi -2{\sin }^{-1}x}{\sqrt{1-x} \left(\sqrt{\pi }+\sqrt{2{\sin }^{-1}x}\right)}=\underset{x\to 1^{-}}{\lim }\frac{2{\cos }^{-1}x}{\sqrt{1-x} \left(\sqrt{\pi }+\sqrt{2{\sin }^{-1}x}\right)}$

Let $K=\underset{x\to 1^{-}}{\lim }\frac{{\cos }^{-1}x}{\sqrt{1-x}}$ and put $x=\cos \theta$, we get

$K=\underset{\theta \to 0}{\lim }\frac{\frac{\theta }{2}.2}{\sqrt{2}.\sin \frac{\theta }{2}}=\underset{\theta \to 0}{\lim }\frac{\sqrt{2}}{{\displaystyle \frac{\sin \frac{\theta }{2}}{\frac{\theta }{2}}}}=\sqrt{2} \left[\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right]$

$\therefore L=\frac{2\sqrt{2}}{2 \sqrt{\pi }}=\sqrt{\frac{2}{\pi }}$.