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Evaluating a limit

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm sadly having trouble evaluating this limit.
    lim n->∞ (n/(n+1)^n

    2. Relevant equations



    3. The attempt at a solution

    I have not done limits for a while. I do remember you have to put it as e to the ln of it and attempt to rearrange it. Just not sure how to start.
     
  2. jcsd
  3. Nov 13, 2012 #2

    haruspex

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    Try writing it as (1+1/n)^n and doing a binomial expansion.
     
  4. Nov 13, 2012 #3
    A binomial expansion?
     
  5. Nov 13, 2012 #4

    Mark44

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    A different approach is to let y = (n/(n + 1))n and then take the ln of both sides. You can then take the limit, which will be amenable to L'Hopital's Rule (with a little manipulation first).

    The idea is that under suitable conditions, the operations of taking the limit and taking the log can be interchanged.
     
  6. Nov 13, 2012 #5
    Won't that make it harder?
     
  7. Nov 13, 2012 #6

    Mark44

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    Harder than what? The limit you have cannot be evaluated directly, as it is in the indeterminate form [1].

    The approach that I suggested gives you a way to evaluate this limit.
     
  8. Nov 13, 2012 #7
    Would I be able to write it in the form lim n-> ∞ e^ln((n/(1+n))^n)

    'cause then I can bring down the n and have n*ln(n/(1+n))
    then i don't know :[. Divide all by n^2?
     
  9. Nov 13, 2012 #8

    Dick

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    Have you used l'Hopital's rule before? You've now got a limit of the form infinity*0. You want write in, say 0/0 form. Like ln(n/(1+n))/(1/n). Now do l'Hopital.
     
  10. Nov 13, 2012 #9
    How can you just take the ln of it though. How can you just divide by 1/n? :(
     
  11. Nov 13, 2012 #10

    Mark44

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    Let y = (n/(n + 1))n

    Take ln of each side:
    ln(y) = ln[(n/(n + 1))n]

    You're not dividing by 1/n. There already is a factor of n. Multiplying by n is the same as dividing by 1/n. Since n is a very large number (and so is a long way from 0), there's no danger of division by 0 in 1/n.

    Your calculus book should have one or more examples of this technique.
     
  12. Nov 13, 2012 #11
    don't you put it the limit as e^ln (n/(n+1))^n ?
     
  13. Nov 13, 2012 #12

    Dick

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    Do you know l'Hopital's rule or not? If you don't then this line is hopeless. Do you know lim n->infinity (1+1/n)^n=e?
     
  14. Nov 13, 2012 #13
    Yesss, I know it. I'm just trying to figure out the notation.
     
  15. Nov 14, 2012 #14

    Dick

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    Ok. Then yes, once you work out lim ln((n/(n+1))^n)=c you find lim (n/(n+1))^n=e^c.
     
  16. Nov 14, 2012 #15

    ehild

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    You know that [tex]\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e[/tex]
    But

    [tex]1+\frac{1}{n}=\frac{1+n}{n}[/tex]

    How is that related to n/(n+1)?

    ehild
     
  17. Nov 14, 2012 #16
    They're the inverse of each other. So this limit would be 1/e?
     
  18. Nov 14, 2012 #17

    Mark44

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    I would say that's a stretch. In the limit that ehild showed, the part in parentheses is (1 + n)/n. If all you do is change that part, and nothing else, it seems a remote possibility that you'll end up with the reciprocal of the limit.

    There have been many good suggestions made in this thread. Why don't pick one and see what you get?
     
  19. Nov 14, 2012 #18
    I've tried using L'Hospital's but idk what I'm doing wrong :(
     
  20. Nov 14, 2012 #19

    Dick

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    You haven't showed what you tried. How did you try? You can do it that way.
     
  21. Nov 14, 2012 #20
    I'm doing :
    Let y = (n/(n + 1))n

    ln(y) = ln[(n/(n + 1))]
    = n ln (n/(n+1))
    = ln (n/(n+1))/(1/n)
     
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