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JEE Main 201816 AprilOnline
Exam
JEE Main
Year
2018
Date
16 April
Mode
Online
Subject
Mathematics
Chapter
Limits

JEE Main 2018 · 16 April · Online · Mathematics · Limits

_ x → 0 (( 27 + x )(1)/(3) - 3)/(9 - ( 27 + x )(2)/(3)) equals

lim⁡x→0(27+x)13−39−(27+x)23\lim_{x \to 0}\dfrac{( 27 + x )\dfrac{1}{3} - 3}{9 - ( 27 + x )\dfrac{2}{3}} equals

Options

  1. A−16- \dfrac{1}{6}
  2. B16\dfrac{1}{6}
  3. C13\dfrac{1}{3}
  4. D−13- \dfrac{1}{3}

Correct answer

A. −16- \dfrac{1}{6}

Step-by-step solution

Given limit is lim⁡x→0(27+x)13−39−(27+x)23(00form)\lim_{x \to 0}\dfrac{\left(27 + x\right)\dfrac{1}{3} - 3}{9 - \left(27 + x\right)\dfrac{2}{3}}\left(\dfrac{0}{0}form\right)

Applying L'Hospital's rule,

⇒lim⁡x→013(27+x)−23−23(27+x)−13=(27)−23−2(27)−13⇒ \lim_{x \to 0}\dfrac{\dfrac{1}{3}{\left(27 + x\right)}^{- \dfrac{2}{3}}}{- \dfrac{2}{3}{\left(27 + x\right)}^{- \dfrac{1}{3}}} = \dfrac{{\left(27\right)}^{- \dfrac{2}{3}}}{- 2{\left(27\right)}^{- \dfrac{1}{3}}}

=−12(27)13=−16= \dfrac{- 1}{2\left(27\right)\dfrac{1}{3}} = - \dfrac{1}{6}.
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