JEE PYQ: Limits Question 33
Question 33 - 2019 (12 Jan Shift 1)
$\lim_{x \to 1^-} \frac{\sqrt{\pi} - \sqrt{2\sin^{-1}x}}{\sqrt{1-x}}$ is equal to:
(1) $\frac{1}{\sqrt{2\pi}}$
(2) $\sqrt{\frac{2}{\pi}}$
(3) $\sqrt{\frac{\pi}{2}}$
(4) $\sqrt{\pi}$
Show Answer
Answer: (2)
Solution
Let $x = 1-h$. $= \lim_{h\to 0}\frac{\sqrt{\pi}-\sqrt{2\sin^{-1}(1-h)}}{\sqrt{h}}$. Rationalising and using $\sin^{-1}(1-h) \approx \frac{\pi}{2} - \sqrt{2h}$: $= 2 \times \frac{1}{\sqrt{\pi}} \times \frac{1}{\sqrt{2}} = \sqrt{\frac{2}{\pi}}$.
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