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Math Help - Find the limit

  1. #1
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    Find the limit

    Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]

    I tried to divide the numerator and denominator by 3^n. Was not successful.

    What should I do next?
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  2. #2
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    Quote Originally Posted by Nichelle14
    Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]

    I tried to divide the numerator and denominator by 3^n. Was not successful.

    What should I do next?
    Did you consider to divide by 3^n+5^n
    Thus,
    \frac{1}{1+\frac{2}{3^n+5^n}}\to 1 as n\to\infty
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  3. #3
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    Also,
    1-\frac{1}{n}\leq \frac{3^n+5^n}{3^n+1+5^n+1} \leq 1+\frac{1}{n}
    Since,
    \lim_{n\to\infty} 1-\frac{1}{n}=\lim_{n\to\infty}1+\frac{1}{n}=1
    Thus,
    \lim_{n\to\infty}\frac{3^n+5^n}{3^n+1+5^n+1}=1 by the squeeze theorem.
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  4. #4
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    Hello, Nichelle14!

    \lim_{n\to\infty}\frac{3^n + 5^n}{3^{n+1} + 5^{n+1}}

    I tried to divide the numerator and denominator by 3^n . . .Was not successful.

    What should I do next?

    Divide top and bottom by 5^{n+1}.

    The numerator is: . \frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;= \frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}

    The denominator is: . \frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1


    Recall that: if |a| < 1, then \lim_{n\to\infty} a^n\:=\:0


    Therefore, the limit is: . \lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3}  {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}<br />
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, Nichelle14!


    Divide top and bottom by 5^{n+1}.

    The numerator is: . \frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;= \frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}

    The denominator is: . \frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1


    Recall that: if |a| < 1, then \lim_{n\to\infty} a^n\:=\:0



    Therefore, the limit is: . \lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3}  {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}<br />
    I think the trick here is that you will divide all the terms by the largest term in the expression(it is useful when the limit tends to infinity)

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    Malay
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