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JEE Main & Advanced Mathematics Functions Question Bank Limits

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{(1+x)}^{5}}-1}{{{(1+x)}^{3}}-1}=\]

    A)                 0

    B)                 1

    C)                 5/3

    D)                 3/5

    Correct Answer: C

    Solution :

                       \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x\,{{[}^{5}}{{C}_{1}}{{+}^{5}}{{C}_{2}}x{{+}^{5}}{{C}_{3}}{{x}^{2}}{{+}^{5}}{{C}_{4}}{{x}^{3}}{{+}^{5}}{{C}_{5}}{{x}^{4}}]}{x\,{{[}^{3}}{{C}_{1}}{{+}^{3}}{{C}_{2}}x{{+}^{3}}{{C}_{3}}{{x}^{2}}]}\] \[=\frac{5}{3}.\]                                 Aliter : Apply L-Hospital?s rule.


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