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Math Help - [SOLVED] Limit we've seen a number of times

  1. #1
    Forum Admin topsquark's Avatar
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    [SOLVED] Limit we've seen a number of times

    I should just look this up, but I can't figure out how to do a search for it. It's cropped up a couple of times, but I don't recall the answer, much less how to find it.

    Anyway, here it is
    \lim_{n \to \infty} \sqrt[n]{n}
    (I'm remembering something about taking the log of the limit perhaps?)

    -Dan
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  2. #2
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    The answer is...

    1


    I will try to remember how to get this!


    A little later...

    Rewrite this as:

    e^(ln(n)/n)

    Then use L'Hostpital on ln(n)/n.

    Lim e^(1/n) = 1
    n→∞
    Last edited by spammanon; February 15th 2008 at 06:17 PM. Reason: Remembered how to do it.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by spammanon View Post
    1


    I will try to remember how to get this!


    Rewrite this as:

    e^(ln(n)/n)

    Then use L'Hostpital on ln(n)/n.

    Lim e^(1/n)
    n→∞
    Aaah! I'd forgotten about that little trick. Thank you. (Yay! I remembered the answer correctly!)

    -Dan
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  4. #4
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    Do it without L'Hopital.

    \ln n = \int_1^n \frac{dx}{x}.

    But, 0\leq \frac{1}{x} \leq \frac{1}{\sqrt{x}} for x\in [1,n] for all n\in \mathbb{Z}^+.

    Thus, 0\leq \int_1^n \frac{dx}{x} \leq \int_1^n \frac{dx}{\sqrt{x}} = 2\sqrt{n} - 2 \leq 2\sqrt{n}.

    Thus, 0 \leq \frac{\ln n}{n} \leq \frac{2\sqrt{n}}{n} = \frac{2}{\sqrt{n}}.

    Now use the Squeeze Theorem.
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