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Homework Help: Limit of ln(n)/ln(n+1) as n->+infinity, very confusing

  1. Oct 26, 2008 #1
    Limit of ln(n)/ln(n+1) as n-->+infinity, very confusing

    can someone help me find the lim as n approaches infinity of


    I used L'HOP so it became (1/n)/(1/n+1) -- as this approaches infinity, it's 0/0, and this confuses me. What am I doing wrong?
  2. jcsd
  3. Oct 26, 2008 #2


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    Homework Helper
    Gold Member

    Re: L'hop

    [itex]\frac{d}{dn} ln(n+1)=\frac{1}{n+1} \neq \frac{1}{n}+1[/itex]

    Does that help ? ;0)
  4. Oct 27, 2008 #3
    Re: L'hop

    first, [tex]\frac{\frac{1}{n}}{\frac{1}{1+n}}=\frac{1+n}{n}=1+\frac{1}{n}[/tex] which approaches to 1, as n approaches to infinity...
    second, I'm assuming n stands for integers. And L'hospitals Rule is not allowed to apply to a sequence.
    Last edited: Oct 27, 2008
  5. Oct 27, 2008 #4


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    Re: L'hop

    L'Hopitals Rule certainly is allowed to apply to a sequence! If f(x) has limit L as x goes to infinity then f(n) has limit L as n goes to infinity. This method is perfectly valid. Of course, you have to do the derivative and the algbra correctly!
  6. Oct 27, 2008 #5
    Re: L'hop

    yeah, I mean this exactly, but applying L'hospitals by brute force is not proper (derivative can be applyed to x but not to n).
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