J & K CET Engineering J and K - CET Engineering Solved Paper-2012

  • question_answer
    The value of \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\int_{0}^{x}{\sin \,({{t}^{2}})\,dt}}{{{x}^{3}}}\]is equal to

    A)  \[0\]              

    B)  \[1\]

    C)  \[3\]               

    D)  None of these

    Correct Answer: D

    Solution :

    \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\int_{0}^{x}{\sin \,({{t}^{2}})dt}}{{{x}^{3}}}\] \[\left( from\,\frac{0}{0} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin ({{x}^{2}})}{3{{x}^{2}}}\]    (L? Hospital?s rule) \[=\frac{1}{3}\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\,\frac{\sin \,({{x}^{2}})}{{{x}^{2}}}\] \[=\frac{1}{3}\times 1=\frac{1}{3}\] \[\left( \because \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin x}{x}=1 \right)\]


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