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Math Help - Limit Question

  1. #1
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    Limit Question

    limit as n-> infinity

    of

    (e^(2n)+e^(n))^(1/n)

    Apparently the answer is e^2 but I'm completely clueless how to solve other than probably using ln. Any help would be nice, thank you!
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  2. #2
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    Quote Originally Posted by negligible91 View Post
    limit as n-> infinity

    of

    (e^(2n)+e^(n))^(1/n)

    Apparently the answer is e^2 but I'm completely clueless how to solve other than probably using ln. Any help would be nice, thank you!


    \displaystyle{\left(e^{2n}+e^n\right)^{1/n}=e^{\frac{1}{n}\ln(e^{2n}+e^n)}} , and now:

    \lim\limits_{n\rightarrow\infty}\frac{\ln(e^{2n}+e  ^n)}{n}=2 , as you can easily check by applying L'Hospital to \lim\limits_{x\rightarrow\infty}\frac{\ln(e^{2x}+e  ^x)}{x}

    Tonio
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    You also can use the sandwich rule...
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  4. #4
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    It is easy to note that e^2  \leqslant \sqrt[n]{{e^{2n}  + e^n }} \leqslant e^2 \sqrt[n]{2}
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  5. #5
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle \ln \{(e^{2n} + e^{n})^{\frac{1}{n}}\}= \frac{1}{n}\ \ln \{e^{2n}\ (1+e^{-n})\} = 2 + \frac{1}{n}\ \ln (1+e^{-n}) (1)

    Now You have to compute the limit of (1) if n tends to infinity...



    Merry Christmas from Italy

    \chi \sigma
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  6. #6
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    Thank you very much everyone, I ended up having trouble using L'Hospital's Rule to simplify to e^2 but chisigma's post helped me find another way to solve the problem as well as use L'Hopital's. Thanks!
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