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

  • question_answer
    Let \[f(x)=\left\{ \begin{matrix}    x\sin \left( \frac{1}{x} \right)+\sin \left( \frac{1}{{{x}^{2}}} \right),x\ne 0  \\    0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0  \\ \end{matrix} \right.\]then \[\underset{x\to \infty }{\mathop{\lim }}\,f(x)\] equals

    A) 0

    B) \[-1/2\]

    C) 1

    D) None of these

    Correct Answer: C

    Solution :

    [c] \[\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,x\sin \left( \frac{1}{x} \right)+\sin \left( \frac{1}{{{x}^{2}}} \right)\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin \left( \frac{1}{x} \right)}{\left( \frac{1}{x} \right)}+\underset{x\to \infty }{\mathop{\lim }}\,\sin \left( \frac{1}{{{x}^{2}}} \right)=1+0=1.\]


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