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U-th root of 1/u as u goes to infinity is 1.

  1. Apr 23, 2009 #1
    Is this true? [tex]n,m\in\mathbb{N}[/tex]
    [tex]\lim_{n\to\infty}\sqrt[n]{\lim_{m\to\infty}\frac{1}{m}}=\lim_{n\to\infty}\lim_{m\to\infty}\sqrt[n]{\frac{1}{m}}=\lim_{u\to\infty}\sqrt{\frac{1}{u}}=1[/tex]
    Thanks
     
  2. jcsd
  3. Apr 24, 2009 #2

    tiny-tim

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    No, that's completely wrong from start to finish (in particular limu -> ∞√u/u is obviously 0).

    What were you trying to do? :confused:
     
  4. Apr 24, 2009 #3
    Re: Limit

    No, u-th root of 1/u as u goes to infinity is 1.
     
  5. Apr 24, 2009 #4

    HallsofIvy

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    Re: Limit

    Gosh, I wish the LaTex were working!
     
  6. Apr 24, 2009 #5

    tiny-tim

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    oh :rolleyes: … did you mean
     
  7. Apr 25, 2009 #6

    arildno

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    Re: Limit

    Well, lukaszh:
    You can't just take two INDEPENDENTLY VARYING parameters, n and m, and convert them into i single parameter!

    To see that let m=n^n.

    Then, the n-th root of m equals n, and the limit of this as n goes to infinity, is infinity.

    Another example. Let m=a^n, where a is some fixed number.

    In this case, the limit equals a!

    What you need to do if you want to prove the convergence of some expresssion as the two parameters to infinity, is that that limit holds, IRRESPECTIVE of how n and m individually increases towards infinity!
     
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