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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{{{x}^{2}}}}-\cos x}{{{\sin }^{2}}x}\]is equal to : JEE Main Online Paper (Held On 10 April 2015)

    A)  3                                            

    B)  \[\frac{3}{2}\]

    C)  2                                            

    D)  \[\frac{5}{4}\]

    Correct Answer: B

    Solution :

                    \[\underset{x\to 0}{\mathop{Lim}}\,\frac{2x{{e}^{{{x}^{2}}}}+\sin x}{2\sin x\cos x}\] \[\underset{x\to 0}{\mathop{Lim}}\,\left( \frac{xx}{\sin x}{{e}^{{{x}^{2}}}}+\frac{1}{2} \right)\frac{1}{\cos x}=1+\frac{1}{2}=\frac{3}{2}\]


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