Prove the following

Question:

$\lim _{x \rightarrow 1-} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}$ ie equal to :

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

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

  3. $\sqrt{\frac{2}{\pi}}$

  4. $\sqrt{\pi}$


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 $x=\cos \theta$

$\lim _{\theta \rightarrow 0^{+}} \frac{2 \theta}{\sqrt{2} \sin \left(\frac{\theta}{2}\right)} \cdot \frac{1}{2 \sqrt{\pi}}=\sqrt{\frac{2}{\pi}}$

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