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Limit as x goes to zero of x^x

  1. Jan 27, 2015 #1

    Nathanael

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    1. The problem statement, all variables and given/known data
    I want to integrate [itex]\int_0^e \ln(x)[/itex] but first, I wondered if it would be divergent. I figured if xx goes to zero as x goes to zero then the integral would diverge (because xln(x)-x would diverge).

    2. The attempt at a solution
    I'm wondering how you could show that this limit (xx as x→0) is 1. It makes intuitive sense that it should approach 1, and it's clear from the graph, but I'm curious as to how someone would find this limit mathematically.
     
  2. jcsd
  3. Jan 27, 2015 #2

    Dick

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    The log of x^x is ln(x)*x. Try and find the limit of that. Write it in such a way that you can use l'Hopital.
     
  4. Jan 27, 2015 #3

    Nathanael

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    Ah, right. I forgot about l'Hopital's idea.

    lim of [itex]\frac{x}{\ln(x)^{-1}}[/itex] as x→0 = lim of [itex]\frac{1}{-(x\ln(x)^2)^{-1}}[/itex] as x→0 = lim of [itex]-(x\ln(x)^2)[/itex] as x→0

    I feel like I could apply this indefinitely with no progress...

    Oh, I see...

    If
    lim of [itex]x\ln(x)[/itex] as x→0 = lim of [itex]-x\ln(x)^2[/itex] as x→0
    then
    lim of [itex]x\ln(x)[/itex] as x→0 = lim of [itex]-x[/itex] as x→0 = 0

    Thanks Dick.

    P.S.
    Sorry if this was hard to read; I don't know how to write the "x→0" under the "lim" in tex. (If someone wants to teach me, it would be appreciated :) )
     
  5. Jan 27, 2015 #4

    Dick

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    I'm not really sure what you are doing there. Try writing x*ln(x) as ln(x)/(1/x) and do l'Hopital.
     
  6. Jan 27, 2015 #5

    Nathanael

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    I thought l'Hopital's idea only applied if the numerator function and denominator function both approach zero? That's why I wrote it as x/(1/ln(x))

    If your way works too that's great (and perhaps a bit simpler) but I arrived at the answer in the previous post.
     
  7. Jan 27, 2015 #6

    Mark44

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    L'Hopital's Rule can also be used when the limit has the form ##[\frac{\infty}{\infty}]##.
     
  8. Jan 27, 2015 #7

    Dick

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    Not really. x*ln(x)^2 isn't any easier as a limit than x*ln(x). I don't know why you just dropped the ln(x)^2.
     
  9. Jan 27, 2015 #8

    Nathanael

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    [itex]\lim\limits_{x\to 0}\big( x\ln(x)\big)=\lim\limits_{x\to 0}\big(-x\ln(x)^2\big)[/itex]

    I divided both sides by -ln(x)

    [itex]\lim\limits_{x\to 0}\big(x\ln(x)\big)=\lim\limits_{x\to 0}(-x)=0[/itex]

    The reason I reversed the order was so it makes more sense reading it from left to right, but now I see that it's confusing. Sorry.
     
  10. Jan 27, 2015 #9

    Dick

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    Now I see. That's an original way to do it. Look ok to me.
     
  11. Jan 28, 2015 #10

    ehild

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    You have shown that ln(xx)=x ln(x) goes to zero at the limit x=0. So xx goes to 1.
    You do not need the limit of xx, as the integral is x(ln(x)-1) . And it is not divergent if x goes to zero.
     
  12. Jan 28, 2015 #11

    Nathanael

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    Right. I was just thinking about xx because I thought perhaps it would be easier to determine the limit of that than the limit of xln(x). It's easier to guess the limit of xx, but to actually calculate the limit we had to go back back to the original xln(x). So xx was indeed irrelevant.
     
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