CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{3.2}^{n+1}}-{{4.5}^{n+1}}}{{{5.2}^{n}}+{{7.5}^{n}}}\] is equal to

    A)  \[\frac{3}{5}\]                                  

    B)  \[-\frac{4}{7}\]

    C)  \[-\frac{20}{7}\]                              

    D)  \[0\]

    Correct Answer: C

    Solution :

    \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{3.2}^{n+1}}-{{4.5}^{n+1}}}{{{5.2}^{n}}+{{7.5}^{n}}}\] \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{5}^{n}}\left( 6.{{\left( \frac{2}{5} \right)}^{n}}-20 \right)}{{{5}^{n}}\left( 5.{{\left( \frac{2}{5} \right)}^{n}}+7 \right)}\] \[=-\frac{20}{7}\]


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