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Limit involving L'Hopital's Rule

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]
    \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
    [/itex]



    3. The attempt at a solution

    [itex]
    y = \displaystyle\lim_{x\rightarrow 0^+} x^{1/x}
    [/itex]

    [itex]
    lny = \displaystyle\lim_{x\rightarrow 0^+} \frac {lnx}{x}
    [/itex]

    This gives an indeterminate form and it's a quotient, so I can apply L'Hopital's Rule:

    [itex]
    \displaystyle\lim_{x\rightarrow 0^+} \frac{1}{x}
    [/itex]

    As x approaches zero, the expression will grow towards infinity...

    [itex]
    lny = ∞
    [/itex]

    [itex]
    e^∞ = y
    [/itex]

    [itex]
    y = ∞
    [/itex]


    I can't see anything wrong with what I'm doing, yet putting the limit into WolframAlpha spits out a 0. What am I doing wrong?
     
  2. jcsd
  3. Sep 22, 2012 #2

    Ray Vickson

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    You do not have an indeterminate quotient, so you should not use l'Hospital's rule. In your case both factors 1/x and log(x) tend to ∞ in magnitude. However, be careful about the sign of log(x) for small x.

    RGV
     
  4. Sep 22, 2012 #3
    Sorry, I don't quite understand.

    As x approaches 0, wouldn't the numerator go towards negative infinity and the denominator towards 0? This gives me (-∞/0) which I don't quite know how to interpret... That isn't considered an indeterminate form?

    Where are my seeing this wrong?
     
  5. Sep 22, 2012 #4

    Ray Vickson

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    I am not allowed to go step-by-step through the argument; you just have to think about it more carefully.

    RGV
     
  6. Sep 22, 2012 #5

    SammyS

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    To be more explicit.

    [itex]\displaystyle
    \ln(y) = \lim_{x\to\,0^+} \frac {\ln(x)}{x}
    [/itex] is not of an indeterminate form because:
    [itex]\displaystyle \ln(x)\to\,-\infty[/itex]

    while [itex]\displaystyle x\to\,0^+\ .[/itex]​
    Use the fact that [itex]\displaystyle \frac{1}{x}\to\,+\infty\ .[/itex]
     
  7. Sep 22, 2012 #6
    Ah, okay. I guess I'm just seeing the problem wrong.

    Thanks to both of you.

    EDIT:

    Ah, I was taking it farther than I needed to. The original expression as it is isn't an indeterminate form and may just be evaluated as is.
     
    Last edited: Sep 22, 2012
  8. Sep 22, 2012 #7

    Ray Vickson

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    I was hoping the OP would figure out that for himself; that is why I did not write it out in detail.

    RGV
     
  9. Sep 23, 2012 #8

    SammyS

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    Ray,

    I apologize for that.

    I just interceded to point out to OP, exactly what line of his post you were commenting on, but I must have been too tired & got carried away.

    Judging by his/her response, I'm still not convinced that OP understands why that expression is not indeterminate.

    SammyS
     
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