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Math Help - Finding limits...

  1. #1
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    Finding limits...

    Hi, I need some help finding the limit of the following....

    \lim_{n \to \infty} \frac{2^n - n}{3^n - n}

    There is also one more i am stuck on,

    \lim_{n \to \infty} \sqrt[n]{2^n + 3^n + 4^n}

    For this one could I use the squeeze theorem?

    Thanks
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  2. #2
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    \lim \frac{2^n-n}{3^n-n} = \lim (\frac{2^n}{3^n-n} - \frac{n}{3^n-n}) = \lim (\frac{2^n}{3^n-n} - \frac{1}{3^n/n-1}) = \lim \frac{2^n}{3^n-n}
    = \lim \frac{2^n}{3^n}\,\frac{1}{1-n/3^n} = \lim \frac{2^n}{3^n} = 0.
    Typically, we would just note that the linear terms ( n) are dominated by the higher-order terms (the exponentials), so they will drop out in the limit. (edit: what Putnam120 said)


    For this one could I use the squeeze theorem
    Yes.
    4 = \sqrt[n]{0+0+4^n} < \sqrt[n]{2^n+3^n+4^n} < \sqrt[n]{4^n+4^n+4^n} < \sqrt[n]3\ 4

    In general \lim_{n\to\infty}(\sum_i |a_i|^n)^{1/n} = \max_i |a_i|.
    Last edited by myrerer; May 15th 2009 at 09:55 PM.
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  3. #3
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    For the first limit, when n is large you have that a^n-n will be almost equal to a^n, for a>1. Thus your limit is now \lim_{n\to\infty}\left(\frac{2}{3}\right)^n=0.
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