JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Morning)

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{3{{x}^{2}}+2}{7{{x}^{2}}+2} \right)}^{\frac{1}{{{x}^{2}}}}}\] is equal to   [JEE MAIN Held On 08-01-2020 Morning]

    A) e

    B) \[\frac{1}{e}\]

    C) \[\frac{1}{{{e}^{2}}}\]

    D) \[{{e}^{2}}\]

    Correct Answer: C

    Solution :

    [c] \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{3{{x}^{2}}+2}{7{{x}^{2}}+2} \right)}^{\frac{1}{{{x}^{2}}}}}\] \[\Rightarrow {{e}^{\underset{x\to 0}{\mathop{\lim }}\,}}\left( \frac{3{{x}^{2}}+2-7{{x}^{2}}-2}{7{{x}^{2}}+2} \right)\frac{1}{{{x}^{2}}}\] \[\Rightarrow {{e}^{\frac{-4}{3}}}={{e}^{-2}}\]


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