JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-{{x}^{2}}}-\sqrt{1+{{x}^{2}}}}{{{x}^{2}}}\] is equal to [MP PET 1999]

    A)                 1

    B)                 ?1

    C)                 ?2

    D)                 0

    Correct Answer: B

    Solution :

                       On rationalising, the given limit                                  \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{(1-{{x}^{2}}-1-{{x}^{2}})}{{{x}^{2}}\,(\sqrt{1-{{x}^{2}}}+\sqrt{1+{{x}^{2}})}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{-2}{\sqrt{1-{{x}^{2}}}+\sqrt{1+{{x}^{2}}}}=\frac{-2}{1+1}=-1\]                


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