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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → 1 ^ - π - 4tan^ - 1 x 1 - x is equal to

The value of lim⁡x→1−π−4tan−1x1−x\lim_{x \to {1}^{-}}\dfrac{\sqrt{\pi} - \sqrt{4tan^{- 1}x}}{\sqrt{1 - x}} is equal to

Options

  1. A2π2\sqrt{\pi}
  2. B12π\dfrac{1}{2\sqrt{\pi}}
  3. C4π4\sqrt{\pi}
  4. D00

Correct answer

D. 00

Step-by-step solution

After rationalization
lim⁡x→1−π−4tan−1x1−x(π+4tan−1x)=12πlim⁡x→1−π−4tan−1x1−x\lim_{x \to {1}^{-}}\dfrac{\pi - 4tan^{- 1}x}{\sqrt{1 - x}\left(\sqrt{\pi} + \sqrt{4tan^{- 1}x}\right)} = \dfrac{1}{2\sqrt{\pi}}\lim_{x \to {1}^{-}}\dfrac{\pi - 4tan^{- 1}x}{\sqrt{1 - x}}
By L-hospital rule, we get,
12πlim⁡x→1−−41+x2−121−x=4πlim⁡x→1−1−x1+x2=0\dfrac{1}{2\sqrt{\pi}}\lim_{x \to {1}^{-}}\dfrac{\dfrac{- 4}{1 + x^{2}}}{\dfrac{- 1}{2\sqrt{1 - x}}} = \dfrac{4}{\sqrt{\pi}}\lim_{x \to {1}^{-}}\dfrac{\sqrt{1 - x}}{1 + x^{2}} = 0
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