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Infinite Limit question (3/x)^2x, proof it goes to 0

  1. Sep 1, 2013 #1
    1. The problem statement, all variables and given/known data
    lim (3/n)^(2n)
    n→ ∞


    2. Relevant equations

    L'hopital's rule: lim F(a)/G(a) is indeterminate form, then the limit can be written as lim F'(a)/G'(a)
    x → a x→ a



    3. The attempt at a solution

    lim e^[ln(3/n)^(2n)] = lim e^(2n * ln(3/n))
    n→ ∞ n → ∞

    2n * ln(3/n) =

    [2 * ln(3/n)] / (n^-1 )

    and applying L.h., [2 * (n/3) * (-3/n^2)]/(-n^-2)

    reduces to, 2 * n, and clearly, plugging in n = infinity is wrong. Where did I go wrong, and what can I do to fix this problem?
     
    Last edited: Sep 1, 2013
  2. jcsd
  3. Sep 1, 2013 #2
    2n * ln(3/n)
    ln(1/x)=?
     
  4. Sep 1, 2013 #3

    vela

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    The Hospital rule doesn't apply here because you don't have an indeterminate form.
     
  5. Sep 1, 2013 #4
    Woops, didn't read the question carefully,,,I art iDiot
     
  6. Sep 1, 2013 #5

    Ray Vickson

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    Please use parentheses: using standard parsing rules for mathematical expressions, what you wrote means
    [tex] \left(\frac{3}{n}\right)^2 n,[/tex]
    but may be you meant
    [tex] \left(\frac{3}{n}\right)^{2 n}.[/tex]
    If you meant the latter, write it as (3/n)^(2n).
     
  7. Sep 1, 2013 #6
    You can see that the the denominator will increase faster than the nominator when ##n>2##. The limit is positive and is bounded by 0. So you just need to find a function which bounds your sequence and converges to 0 also when the limit goes to infinity. In this case I suggest having a function in form of geometric sequence. If that's so then you can prove that this series reaches 0.
     
  8. Sep 2, 2013 #7

    vanhees71

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    It's pretty easy to show that the sequence goes to 0 for [itex]n \rightarrow \infty[/itex], if you consider the logarithm of the expression.
     
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