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JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{\int\limits_{0}^{x}{{{\tan }^{-1}}t\,dt}}{\sqrt{{{x}^{2}}+1}}\] has the value

    A)  \[\frac{\pi }{2}\]                                   

    B)  0

    C)  1                                

    D)  \[\pi \]

    Correct Answer: A

    Solution :

    Use L?Hosptial?s rule \[\underset{x\to \infty }{\mathop{Lim}}\,\,\frac{({{\tan }^{-1}}x)\sqrt{{{x}^{2}}+1}}{x}=\frac{\pi }{2}\,\underset{x\to \infty }{\mathop{Lim}}\,\frac{\sqrt{{{x}^{2}}+1}}{x}=\frac{\pi }{2}\]


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