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How do I show that x^x->1 as x->0?

  1. Mar 16, 2006 #1
    How do I show that x^x->1 as x->0?
     
  2. jcsd
  3. Mar 16, 2006 #2

    shmoe

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    Try using log's to convert the exponentiation into a product.
     
  4. Mar 16, 2006 #3
    I can't find a lower bound for xlog(x).
     
  5. Mar 16, 2006 #4
    use l'hopital's rule.

    if you have lim x log x then x log x is the same as [log x / (1 / x)]
     
  6. Mar 16, 2006 #5
    Got it. Thanks.
     
  7. Mar 16, 2006 #6

    shmoe

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    You can also change it to what may be a more familiar limit by setting x=1/y.
     
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