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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2020 · 8 January · Morning Shift · Mathematics · Limits

_ x → 0 ( 3 x ^ 2 + 2 7 x ^ 2 + 2 ) 1 x ^ 2 is equal to

lim⁡x→0(3x2+27x2+2)1x2\lim_{x \to 0}\left(\dfrac{3{x}^{2} + 2}{7{x}^{2} + 2}\right)\dfrac{1}{{x}^{2}} is equal to

Options

  1. A1e\dfrac{1}{e}
  2. B1e2\dfrac{1}{{e}^{2}}
  3. Ce2{e}^{2}
  4. Dee

Correct answer

B. 1e2\dfrac{1}{{e}^{2}}

Step-by-step solution

Let L=lim⁡x→0(3x2+27x2+2)1x2L = \lim_{x \to 0}\left(\dfrac{3{x}^{2} + 2}{7{x}^{2} + 2}\right)\dfrac{1}{{x}^{2}}

=elim⁡x→01x2{3x2+27x2+2−1}= {e}^{\lim_{x \to 0}\dfrac{1}{{x}^{2}}\left\{\dfrac{3{x}^{2} + 2}{7{x}^{2} + 2} - 1\right\}}

=elim⁡x→0{−4x27x2+2}= {e}^{\lim_{x \to 0}\left\{\dfrac{- 4{x}^{2}}{7{x}^{2} + 2}\right\}}

=e−42= e\dfrac{- 4}{2}

=e−2= {e}^{- 2}

=1e2= \dfrac{1}{{e}^{2}}

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