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Limit with multiple zeros

  1. Nov 21, 2008 #1
    I have been working with a limit for a while now but cannot for the life of me seem to solve it. Any ideas:

    lim[x appr. 0] of (x^x)/((e^x)-1)

    I've tried turning x^x into e^(x ln(x)), but the root of my problem is that I'm unsure of whether or not I can use L'Hopital's because technically, (0^0)/0 is not necessarily of the form 0/0.
  2. jcsd
  3. Nov 21, 2008 #2


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    e^x - 1 ~ x. Therefore your expression is ~ x^(1+x) -> 0^1 = 0.
  4. Nov 21, 2008 #3
    You make a good point, but, assuming that is correct, wouldn't it go to:

    x^(x-1) >>> oo
  5. Nov 21, 2008 #4


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    That is right.
    This is straight forward
  6. Nov 22, 2008 #5


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    You're right. I misread your original expression - I missed the / .
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