Medium MCQ +4 / -1 PYQ · JEE Mains 2020

$$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ is equal to

  1. A e
  2. B e<sup>2</sup>
  3. C ${1 \over {{e^2}}}$ Correct answer
  4. D ${1 \over e}$

Solution

Given $$\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$$ <br><br>Putting x = 0 we get 1<sup>$\infty$</sup> form. <br><br>$\therefore$ $${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{3{x^2} + 2} \over {7{x^2} + 2}} - 1} \right]}}$$ <br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{ - 4{x^2}} \over {7{x^2} + 2}}} \right]}}$$ <br><br>= e<sup>-4/2</sup> <br><br>= e<sup>-2</sup> = ${1 \over {{e^2}}}$

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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