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Differential equation

  1. Nov 1, 2012 #1

    fxo

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    1. The problem statement, all variables and given/known data

    Use L'Hospitals rule to show that [itex]lim x->0 x^a ln(x) = 0 [/itex]

    I don't know how to solve this. I guess the first thing to do is to transform it in some way so that one can use L'Hospitals rule, but I don't know how.

    Thank you!

    EDIT: a>0

    It's not a differential equation as the headline says, my bad.
     
    Last edited: Nov 1, 2012
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  3. Nov 1, 2012 #2

    LCKurtz

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    Is there something else you haven't told us? Like ##a>0##? Write it like this:$$
    \lim_{x\rightarrow 0}\frac {\ln x}{\frac 1 {x^a}}$$
     
  4. Nov 1, 2012 #3

    fxo

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    Oh sorry you're right I totally missed that a>0. I get it then.

    Can I ask a follow-up question based on this one?

    By setting x=ln(t) in the equation above(the one I asked for help on first) show that:

    $$
    \lim_{x\rightarrow -∞} |x|^a e^x = 0$$

    This one really bugs my mind I just don't get it.
     
  5. Nov 1, 2012 #4

    LCKurtz

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    You should note that in the original limit, it is really ##x\rightarrow 0^+##. What happens if you substitute ##x=\ln t## in the original problem?
     
  6. Nov 1, 2012 #5

    fxo

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    $$
    \lim_{ln(t)\rightarrow 0}\frac {\ln (ln(t))}{\frac 1 {ln(t)^a}}$$

    So I get $$
    ln(ln(t))ln^a(t)$$

    But that expression don't tell me much all I know is that it's root should be e.
     
  7. Nov 1, 2012 #6

    LCKurtz

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    Woops, I didn't mean that. I meant to try ##t=\ln x## which is ##x=e^t##. I expect that is a typo in your text.
     
  8. Nov 1, 2012 #7

    fxo

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    ##\lim_{e^t\rightarrow 0}\frac {\ln (e^t}{\frac 1 {e^t^a}}##

    I don't really get it what's the connection to the new limit, do they want me to change the e^x to x=ln(t) and get |x|^a * t instead?
     
  9. Nov 1, 2012 #8

    LCKurtz

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    C'mon, do the algebra simplification. And express it as$$
    \lim_{t\rightarrow\, ?}....$$Then you might see a connection to your new problem.
     
  10. Nov 2, 2012 #9

    fxo

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    By setting ##x=e^t## in the first one I got


    $$
    \lim_{t\rightarrow\, 0} te^t^a $$
    I'm starting to see that it's close to the form in b but not exactly, should I use L'Hospitals rule from here to show that as this goes to infinity the limit goes to zero?
     
  11. Nov 2, 2012 #10

    LCKurtz

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    Tex won't accept e^t^a with no parentheses. Anyway, do you consider that to be simplified? It is more properly written ##(e^t)^a##, but still not simplified until you write it as ##e^{at}##. And you have ##t\rightarrow 0##. If ##x=e^t## and ##x\rightarrow 0^+##, does ##t\rightarrow 0##? What should it be?

    That being said, you are correct that you can massage it into ##|x|e^x## form but it still isn't in the ##|x|^ae^x## form requested. So I'm afraid my suggestion to correct what I think is a misprint isn't going to finish it. To tell the truth, I don't see why your author wants you to do it that way anyway. It is easy to work it directly. Letting ##x = -t## we have$$
    \lim_{x\rightarrow {-\infty}}|x|^ae^x=\lim_{t\rightarrow \infty}t^ae^{-t}
    =\lim_{t\rightarrow \infty}\frac{t^a}{e^{t}}$$which you can now do with repeating L'Hospitals rule until the exponent on ##t## goes negative.
     
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