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Limit problem

  1. Apr 22, 2008 #1
    1. The problem statement, all variables and given/known data

    find the limit as n->oo of n(ln(n+10)-ln(n))

    2. Relevant equations

    3. The attempt at a solution

    the answer is 10 this is also part of my exam review I am pretty lost on this one I tried using log rules to combine the terms tried getting it in form for L'H but I don't know what to do.
  2. jcsd
  3. Apr 22, 2008 #2


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    What did you try?
  4. Apr 22, 2008 #3
    I think I just found the answer...I combined the ln terms so ln((n+10)/n)^n = L then I exponentiated both sides to give ((n+10)/n)^n= e^L then The limit of (1 + 10/n)^n = e^10 = e^L so taking the ln of both sides gives 10 = L (I know the limit of (1 + 10/n)^n = e^10 because it is a given formula on my formula sheet)
  5. Apr 22, 2008 #4


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    That's one way. To use log rules try ln(n+10)-ln(n)=ln((n+10)/n)=ln(1+10/n). Now write it as ln(1+10/n)/(1/n) which is 0/0 and do l'Hopital.
  6. Apr 22, 2008 #5
    I keep getting the wrong answer with this method I do L'H from where you said and I get (n/(1+10)*(-1(1+10)/n^2)*-n^2 ...
    Last edited: Apr 22, 2008
  7. Apr 22, 2008 #6
    finally got it thanks
  8. Apr 23, 2008 #7
    Even if you got it, I can point out another method. Combine the two logarithms to get n*ln(1+10/n), then write ln as an infinite sum (ln(1+x)=x-x^2/2+x^3/3-... if -1<x<1, which obviously is true for n->oo), and the solution comes immediately.
  9. Apr 24, 2008 #8
  10. Apr 24, 2008 #9
    the lim.solution:

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