JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,{{(1-ax)}^{\frac{1}{x}}}=\] [Karnataka CET 2003]

    A)                 e

    B)                 \[{{e}^{-a}}\]

    C)                 1

    D)                 \[{{e}^{a}}\]

    Correct Answer: B

    Solution :

                    \[\underset{x\to 0}{\mathop{\lim }}\,\,{{[1+(-a)\,x]}^{1/x}}={{e}^{-a}}\].


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