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Limit of function with log

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the limit of
    f(x) = x^2 ln(1+1/x) - x

    x-> ∞




    3. The attempt at a solution

    Not sure where to start, but I take the derivative

    f'(x) = [itex]2x*ln(1+1/x) +x^2 \frac{ -x^{-2}}{1+1/x} -1 [/itex]
    [itex]2x*ln(1+1/x) - \frac{1}{1+1/x} -1 [/itex]

    the second term goes to -1 as x->∞ ant the last term is always -1. Can I say somethong about the last term and will it help me?
     
  2. jcsd
  3. Nov 16, 2011 #2
    Got it.

    f(x) = x^2 ln(1+1/x) - x

    = [itex]\frac{ln(1+1/x) - (1/x)}{1/x^2}[/itex]

    u = 1/x

    f(u) = (ln(1+u) -u)/u^2

    which limit is -1/2 when u->0

    The limit is found by using l hopital twice
     
    Last edited: Nov 16, 2011
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