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Thread: x^x

  1. March 16th 2006, 12:11 PM #1
    Treadstone 71
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    x^x

    How do I show that x^x->1 as x->0?
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  2. March 16th 2006, 12:19 PM #2
    rgep
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    You could try showing that x \log x \rightarrow 0 as x \rightarrow 0. If the limit exists it must be zero: consider x = 2^{-n}.
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  3. March 16th 2006, 12:49 PM #3
    Treadstone 71
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    Got it. Used l'Hopitals
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