Limit Continuity and Differentiability 2 Question 2
2. $\lim _{x \rightarrow 1-} \frac{\sqrt{\pi}-\sqrt{2 \sin ^{-1} x}}{\sqrt{1-x}}$ is equal to
(a) $\sqrt{\frac{\pi}{2}}$
(b) $\sqrt{\frac{2}{\pi}}$
(c) $\sqrt{\pi}$
(d) $\frac{1}{\sqrt{2 \pi}}$
(2019 Main, 12 Jan II)
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Answer:
Correct Answer: 2. (c)
Solution:
- $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int _2^{4} f(t) d t}{x^{2}-\frac{\pi^{2}}{16}}$
$=\lim _{x \rightarrow \pi / 4} \frac{f\left(\sec ^{2} x\right) 2 \sec x \sec x \tan x}{2 x}$
$=\frac{2 f(2)}{\pi / 4}=\frac{8}{\pi} f(2)$
[using L’ Hospital’s rule]
