`
`
Question:
$\lim _{x \rightarrow 1-} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}$ ie equal to :
Correct Option: , 3
Solution:
$\lim _{x \rightarrow 1^{-}} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}} \times \frac{\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}}{\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}}$
$\lim _{x \rightarrow 1^{-}} \frac{2\left(\frac{\pi}{2}-\sin ^{-1} x\right)}{\sqrt{1-x} \cdot\left(\sqrt{\pi}+\sqrt{2 \sin ^{-1} x}\right)}$
$\lim _{x \rightarrow 1^{-}} \frac{2 \cos ^{-1} x}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{\pi}}$
Put $x=\cos \theta$
$\lim _{\theta \rightarrow 0^{+}} \frac{2 \theta}{\sqrt{2} \sin \left(\frac{\theta}{2}\right)} \cdot \frac{1}{2 \sqrt{\pi}}=\sqrt{\frac{2}{\pi}}$
.jpg)