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Help with limit please.

  • Thread starter laker88116
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  • #1
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[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?
 
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Answers and Replies

  • #2
VietDao29
Homework Helper
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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,
 
  • #3
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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.
 
  • #4
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Because [tex] (x^x)^2 = x^{2x}[/tex] and it would seem silly to write it that way.
 
  • #5
57
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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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