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JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}3x}{{{x}^{2}}}=\] [Roorkee 1982; DCE 1999]

    A)                 6

    B)                 9

    C)                 18

    D)                 3

    Correct Answer: C

    Solution :

                    \[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{2\times 9\,{{\sin }^{2}}3x}{{{(3x)}^{2}}}=18\]


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