SATHEE: Limit Continuity and Differentiability 2 Question 2

Limit Continuity and Differentiability 2 Question 2

2. $lim_{x \to 1 -} \frac{\sqrt{π} - \sqrt{2 \sin^{- 1} x}}{\sqrt{1 - x}}$ is equal to

(a) $\sqrt{\frac{π}{2}}$

(b) $\sqrt{\frac{2}{π}}$

(d) $\frac{1}{\sqrt{2 π}}$

(2019 Main, 12 Jan II)

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Answer:

Correct Answer: 2. (c)

Solution:

$lim_{x \to \frac{π}{4}} \frac{\int_{2}^{4} f (t) d t}{x^{2} - \frac{π^{2}}{16}}$

$= lim_{x \to π / 4} \frac{f (\sec^{2} x) 2 \sec x \sec x \tan x}{2 x}$

$= \frac{2 f (2)}{π / 4} = \frac{8}{π} f (2)$

[using L’ Hospital’s rule]

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Limit Continuity and Differentiability 2 Question 2

2. limx→1−π−2sin−1⁡x1−x is equal to

Answer:

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2. $lim_{x \to 1 -} \frac{\sqrt{π} - \sqrt{2 \sin^{- 1} x}}{\sqrt{1 - x}}$ is equal to