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Help with limit please.

  1. Aug 23, 2005 #1
    [tex] \lim x \rightarrow 0+[/tex] [tex] x^x^2 [/tex]. This should read x^x^2 if that isn't clear. I am not sure where to start. I know that I need to use L'Hopitals Rule and I know that I need to get this in a form of [itex] \frac{f(x)}{g(x)}[/itex]. Any suggestions on what else to do?
     
    Last edited: Aug 23, 2005
  2. jcsd
  3. Aug 23, 2005 #2

    VietDao29

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    Homework Helper

    Hmm, if you take logs of both sides, what could you get?
    Let [tex]y = \lim_{x \rightarrow 0 ^ +} x^{(x ^ 2)}[/tex]
    So:
    [tex]\ln y = \lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x = z[/tex]
    Now, to find:[tex]\lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x[/tex], you can use L'Hopital rule. Rewrite it as:
    [tex]z = \lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x = \lim_{x \rightarrow 0 ^ +} \frac{\ln x}{\frac{1}{x ^ 2}}[/tex]
    So [tex]y = e ^ z[/tex].
    Viet Dao,
     
  4. Aug 23, 2005 #3
    But how do you know that its isn't [tex](x^x)^2[/tex]? I understand what you are saying though and it makes perfect sense, I just wasn't sure where to break it up.
     
  5. Aug 23, 2005 #4
    Because [tex] (x^x)^2 = x^{2x}[/tex] and it would seem silly to write it that way.
     
  6. Aug 23, 2005 #5
    Ok, well I understand now I was able to interpret it after VietDao29 did most of the work, but anyways, thanks for the help I see it now.
     
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