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L'Hôpital's rule

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the limit as x->0+ of (ln(x))^x

    *The answer is 1*

    2. Relevant equations

    l'Hôpital's rule

    3. The attempt at a solution

    lim (ln(x))^x = 0^0

    I took the ln of that quantity to bring down the x

    lim = x*ln(ln(x))

    lim = ln(ln(x)) / (1/x)

    Then I used l'Hôpital's rule


    = 1/(x^3*ln(x))

    I got stuck here. If I plug in zero I get 0^3 and a undefined answer in the denominator. Do I have to do l'Hôpital's rule on the bottom again?
    Last edited: Nov 15, 2008
  2. jcsd
  3. Nov 16, 2008 #2


    Staff: Mentor

    It seems that you edited this problem while I was composing an answer. The first problem I saw was lim (-ln(x))^x, as x approached 0+.
    A little while later, the minus sign in front of ln(x) was gone. Here's my work for the original problem. For the edited problem, no limit exists.

    [tex]Let y = (-ln x)^x[/tex]
    [tex]ln y = x ln(-ln(x)) = \frac{ln(-ln(x))}{\frac{1}{x}}[/tex]

    [tex]lim_{x \rightarrow 0^+} ln y = lim_{x \rightarrow 0^+} \frac{ln(-ln(x))}{\frac{1}{x}} [\frac{\infty}{\infty}][/tex]

    [tex]= lim_{x \rightarrow 0^+} \frac{\frac{1}{-ln(x)}\frac{-1}{x}}{\frac{-1}{x^2}} [/tex]
    [tex]= lim_{x \rightarrow 0^+} \frac{-x}{ln(x)} = 0[/tex]
    Since [tex]lim_{x \rightarrow 0^+} ln y = 0[/tex]
    Then [tex]ln(lim_{x \rightarrow 0^+} y) = 0[/tex]
    Or [tex]lim_{x \rightarrow 0^+} (-ln(x))^x = 1[/tex]
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