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Math Help - Evaluate a Limit

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    Evaluate a Limit

    Find the lim as x tends to 0+ of (ln x)^x.
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    Quote Originally Posted by h2osprey View Post
    Find the lim as x tends to 0+ of (ln x)^x.
    Is this supposed to be in the complex plane? because otherwise this doesn't make any sense (your x \rightarrow 0^+ puzzles me though) since (\ln x )^x is not even defined (in \mathbb{R} ) in (0,1)
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    Quote Originally Posted by Jose27 View Post
    Is this supposed to be in the complex plane? because otherwise this doesn't make any sense (your x \rightarrow 0^+ puzzles me though) since (\ln x )^x is not even defined (in \mathbb{R} ) in (0,1)
    I doubt so. It's not defined at all points, but not undefined in the whole interval - consider x = 1 / (2k+1), where k is any positive integer.
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    Quote Originally Posted by h2osprey View Post
    I doubt so. It's not defined at all points, but not undefined in the whole interval - consider x = 1 / (2k+1), where k is any positive integer.
    The number y^x is only defined for positive y, because y^x=e^{x\ln y} and \ln x >0 iff x \in (1, \infty) therefore (\ln x )^x is only defined in (1, \infty)
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    Quote Originally Posted by Jose27 View Post
    The number y^x is only defined for positive y, because y^x=e^{x\ln y} and \ln x >0 iff x \in (1, \infty) therefore (\ln x )^x is only defined in (1, \infty)
    You made a mistake -- it's ln y, not ln x.
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    Quote Originally Posted by h2osprey View Post
    You made a mistake -- it's ln y, not ln x.
    No, that logarithm refers to the one in your problem.
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    Quote Originally Posted by Jose27 View Post
    No, that logarithm refers to the one in your problem.
    Okay to illustrate my point:

    Try using a calculator to evaluate (ln (1/3))^ (1/3) and see what you get.
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    Quote Originally Posted by Jose27 View Post
    The number y^x is only defined for positive y, because y^x=e^{x\ln y} and \ln x >0 iff x \in (1, \infty) therefore (\ln x )^x is only defined in (1, \infty)
    And the mistake here is that you brought down the power x. y can be negative if y^x is positive as ln (y^x) would be defined.
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    Quote Originally Posted by h2osprey View Post
    Okay to illustrate my point:

    Try using a calculator to evaluate (ln (1/3))^ (1/3) and see what you get.
    Okay, I think I get it: What you want is \lim_{k \rightarrow \infty } \ \ln ( \frac{1}{2k+1} )^{ \frac{1}{2k+1} } not using the definition of x^x for x \in \mathbb{R} ^+ but the fact that negative numbers have odd roots. Right?
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    Quote Originally Posted by h2osprey View Post
    Find the lim as x tends to 0+ of (ln x)^x.
    (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}. So you should first consider \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}} which has the indeterminant form \frac{\infty}{\infty}. Using l'Hopitals rule is an obvious thing to do.
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    \lim_{k \rightarrow \infty } \ \ln ( \frac{1}{2k+1} )^{ \frac{1}{2k+1} }  = \lim_{k \rightarrow \infty } \ (-\ln ( 2k+1 ))^{ \frac{1}{2k+1} }  = -\lim_{k \rightarrow \infty } \ (\ln ( 2k+1 ))^{ \frac{1}{2k+1} }  = -\lim_{x \rightarrow \infty } \ (\ln ( x ))^{ \frac{1}{x} }  = -\lim_{x \rightarrow \infty} \ e^{  \frac{1}{x} \ln ( \ln x)}  = -e^{ \lim_{x \rightarrow \infty} \ \frac{\ln ( \ln x)}{x}  }. Now use L'Hopital.

    (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}. So you should first consider \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}} which has the indeterminant form \frac{\infty}{\infty}. Using l'Hopitals rule is an obvious thing to do.
    The problem with that is that the limit you get is 1, when the actual limit is -1
    Last edited by Jose27; October 9th 2009 at 09:19 PM.
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    i agree with Jose27: \lim_{x\to0+}(\ln x)^x does not exist because there is no right neighbourhood of x = 0 which is a subset of the domain (\ln x)^x.
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    The problem is, Wolfram Alpha is calculating the limit in the complex plane with the definition I first gave, but by what I understood the limit that was asked ignores this and uses only the fact negative numbers have odd roots (ie. we want the limit of a sequence, not the limit of the function)
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    Quote Originally Posted by mr fantastic View Post
    (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}. So you should first consider \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}} which has the indeterminant form \frac{\infty}{\infty}. Using l'Hopitals rule is an obvious thing to do.
    I'm confused, isn't  \lim_{x \rightarrow 0^+}\ln (\ln x) undefined?
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