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L'Hopital's Rule

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Use L'Hopital's rule to evaluate the limit:

    lim (x/(x+1))^x

    2. Relevant equations

    3. The attempt at a solution

    I put it into a logarithm first to make it the limit of xln(x/(x+1)) then I took the derivative and got [(1/x)-(1/(x+1))]/[x^-2]
    but its still indeterminate and after taking the 2nd derivative it was as well
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    my guess would be you have to use some log properties like log(a/b) = log(a) - log(b)
  4. Nov 12, 2007 #3
    i did do that to get to the derivative but the derivatives didn't help
  5. Nov 12, 2007 #4
    You forgot to use the power rule since there are 2 functions

    This is what I got (could be wrong)

    = x(logx - log (x+1)
    = logx - log(x+1) + x((1/x)-(1/x-1)) <---simplify
    = logx - log(x+1) - (1/x-1)

    I believe you need to take the second derivative and you should get the limit to be 0 (it's been about a year since I did diffrential calc)
  6. Nov 12, 2007 #5
    Why use L'Hopital? Just look at its reciprocal function. Trivial.
  7. Nov 12, 2007 #6
    To get more experience using it?
  8. Nov 13, 2007 #7


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    Staff Emeritus
    Science Advisor

    Why the derivative? L'Hospital's rule usually involves taking two derivatives separately! Write the logarithm as
    [tex]\frac{ln(x)- ln(x+1)}{x^{-1}}[/tex]
    and apply L'Hopital's rule to that.
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