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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    Let\[p=\underset{x\to 0+}{\mathop{\lim }}\,{{\left( 1+{{\tan }^{2}}\sqrt{x} \right)}^{\frac{1}{2x}}}\] then log p is equal to :- [JEE Main Solved Paper-2016 ]

    A) \[\frac{1}{4}\]                                   

    B)  2

    C) 1                                             

    D) \[\frac{1}{2}\]

    Correct Answer: D

    Solution :

                    \[p={{e}^{\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{2}{{\left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right)}^{2}}}}=\sqrt{e}\]             \[\log p=\frac{1}{2}\]


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