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X ln(x)

  1. Oct 10, 2008 #1
    I am trying to prove that the upper +limit of x^x, when x->0 converges to 1.

    So I started by converting x^x to e^(x ln(x)). I know that this eliminates the domain: x <= 0, but I still believe that I can still continue on.

    So here I tried to constrain the limit: x ln(x) (i.e. x->0, x ln(x) -> 0; which is where I failed. Although I can show via the L'hopital's Rule that it is true, I struggle to show it via the Epsilon Delta Definition.

    I know that x^x is undefined at 0, but I still want to show that the curve converges towards 1.
     
  2. jcsd
  3. Oct 10, 2008 #2

    arildno

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    Why not show that L'Hopitul's rule can be proven in general with the epsilon/delta-definition?
     
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