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Finding the Limit! Help!

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the limit of: lim as n--> ∞ of [ln n / ln (n+1)]^n


    2. Relevant equations



    3. The attempt at a solution
    I used the Lopitals rule to find the limit of ln/ln(n+1) and it equals to 1.
    However what do i do after there?
    AND I know its bounded by 0< <1
    squeeze theorem..
    BUT IM not sure what to do next
     
  2. jcsd
  3. Apr 12, 2012 #2

    SammyS

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    Hello juliusoh. Welcome to PF !

    Try to find the limit of the log of your expression.

    [itex]\displaystyle \lim_{n\to\infty} \ln\left(\left(\frac{\ln(n)}{\ln(n+1)}\right)^n\ \right)[/itex]
     
  4. Apr 13, 2012 #3
    Wow its too complicated.. i keep getting stuck....
     
  5. Apr 13, 2012 #4
    This is a good problem to get you acquainted with l'Hopital's rule :wink:

    Is this how you started it? I haven't worked it out, but this is how I would rewrite it before l'Hopital's rule:

    [tex]\ln \left(\left(\frac{\ln x}{\ln(x+1)}\right)^x\right) = x\ln\left(\frac{\ln x}{\ln(x+1)}\right) = \frac{\ln\left(\frac{\ln x}{\ln(x+1)}\right)}{\frac{1}{x}}[/tex]
     
  6. Apr 14, 2012 #5

    Dick

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    If you are just interesting in finding the limit and less interested in a rigorous proof, you could do it the physicists way. Use approximations like log(1+x)~x and 1/(1+x)~1-x where x<<1 and the '~' means I've left out higher order terms in the taylor expansion. Start by writing log(n+1)=log(n*(1+1/n))=log(n)+log(1+1/n)~log(n)+1/n.
     
    Last edited: Apr 14, 2012
  7. Apr 14, 2012 #6
    I see!
    Thanks BohRok..
    I dont know how to do the physics way, i havent learned that yet. Im just a freshman in college.
    So derivative of ln(ln x/ln(x+1).... how do you do that lol.
     
  8. Apr 14, 2012 #7
    d/dx ln(f(x)) = f'(x)/f(x), then let f(x) = ln x/ln(x+1). Be careful when using the quotient rule to find f'(x)!
     
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