JEE Main & Advanced Mathematics Differentiation Question Bank Self Evaluation Test - Limits and Derivatives

  • question_answer
    \[\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{{{5}^{n+1}}+{{3}^{n}}-{{2}^{2n}}}{{{5}^{n}}+{{2}^{n}}+{{3}^{2n+3}}}\] is equal to

    A) 5

    B) 3

    C) 1

    D) 0

    Correct Answer: D

    Solution :

    [d] \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{5}^{n+1}}+{{3}^{n}}-{{2}^{2n}}}{{{5}^{n}}+{{2}^{n}}+{{3}^{2n+3}}}=\underset{n\,\to \,\infty }{\mathop{\lim }}\,\frac{{{5.5}^{n}}+{{3}^{n}}-{{4}^{n}}}{{{5}^{n}}+{{2}^{n}}+{{27.9}^{n}}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{5.\frac{{{5}^{n}}}{{{9}^{n}}}+\frac{{{3}^{n}}}{{{9}^{n}}}-\frac{{{4}^{n}}}{{{9}^{n}}}}{\frac{{{5}^{n}}}{{{9}^{n}}}+\frac{{{2}^{n}}}{{{9}^{n}}}+27}=\frac{0+0-0}{0+0+27}=0.\]


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