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JEE Main & Advanced Mathematics Differentiation Question Bank Self Evaluation Test - Limits and Derivatives

  • question_answer
    Let \[f:R\to R\] be such that \[f(1)=3\] and \[f'(1)=6.\] Then \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{f(1+x)}{f(1)} \right)}^{1/x}}\] equals

    A) 1

    B) \[{{e}^{1/2}}\]

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

    D) \[{{e}^{3}}\]

    Correct Answer: C

    Solution :

    [c] Given that \[f:R\to R\] such that \[f(1)=3\,\,and\,\,f'(1)=6\] Then \[\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{f(1+x)}{f(1)} \right]}^{1/x}}\] \[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{x}[logf(1+x)-logf(1)]}}\] \[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{f(1+x)}f'(1+x)}{1}}}={{e}^{\frac{f'(1)}{f(1)}}}={{e}^{6/3}}={{e}^{2}}\] [Using L Hospital rule]


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