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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 → 0 1 + sin x - cos x + (1 - x) x · ^ 2 x & is

The value of lim⁡x→01+sinx−cosx+ln⁡(1−x)x⋅tan⁡2x\lim_{x \to 0}\dfrac{1 + sin x - cos x + \ln\left(1 - x\right)}{x \cdot \tan^{2}x}& is

Options

  1. A−12- \dfrac{1}{2}
  2. B−13- \dfrac{1}{3}
  3. C12\dfrac{1}{2}
  4. D14\dfrac{1}{4}

Correct answer

A. −12- \dfrac{1}{2}

Step-by-step solution

lim⁡x→01+sinx−cosx+ℓn(1−x)x⋅tan⁡2x\lim_{x \to 0}\dfrac{1 + sin x - cos x + ℓn\left(1 - x\right)}{x \cdot \tan^{2}x}

=lim⁡x→01+sinx−cosx+ℓn(1−x)x3(00)[∵lim⁡x→0xtan⁡x=1]= \lim_{x \to 0}\dfrac{1 + sin x - cos x + ℓn\left(1 - x\right)}{x3}\left(\dfrac{0}{0}\right)[ ∵ \lim_{x \to 0} \dfrac{x}{\tan x} = 1 ]

=lim⁡x→0cosx+sinx−11−x3x2(00)[usingL′hospitalrule]<br/><br/>= \lim_{x \to 0}\dfrac{cos x + sin x - \dfrac{1}{1 - x}}{3x^{2}} \left(\dfrac{0}{0}\right) [using L'hospital rule]<br /><br />= \lim_{x \to 0}−sinx+\cosx−1(1−x)26x(00)[UsingL′hospitalruleagain]<br/><br/>\dfrac{- sin x + \text\cosx - \dfrac{1}{( 1 - x{)}^{2}}}{6x} \left(\dfrac{0}{0}\right) [Using L'hospital rule again]<br /><br />= \lim_{x \to 0}\dfrac{- cos x - \text\sinx - \dfrac{2}{( 1 - x{)}^{3}}}{6} [Using L'hospital rule again]<br /><br />=\dfrac{- 1 - 2}{6} = - \dfrac{1}{2}
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