JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    The value of  \[\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{1-{{n}^{2}}}{\sum n}\] will be [UPSEAT 1999]

    A)                 ? 2

    B)                 ? 1

    C)                 2

    D)                 1

    Correct Answer: A

    Solution :

                       \[\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1-{{n}^{2}}}{\Sigma n}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{(1-n)(1+n)}{\frac{1}{2}n(n+1)}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{2\,(1-n)}{n}\]                                       \[=\underset{n\to \infty }{\mathop{\lim }}\,2\,\left( \frac{1}{n}-1 \right)\]\[=2(0-1)=-2\].


You need to login to perform this action.
You will be redirected in 3 sec spinner