# Evaluate a Limit

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• Oct 9th 2009, 05:29 PM
h2osprey
Evaluate a Limit
Find the lim as x tends to 0+ of (ln x)^x.
• Oct 9th 2009, 06:43 PM
Jose27
Quote:

Originally Posted by h2osprey
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 $\displaystyle x \rightarrow 0^+$ puzzles me though) since $\displaystyle (\ln x )^x$ is not even defined (in $\displaystyle \mathbb{R}$ ) in $\displaystyle (0,1)$
• Oct 9th 2009, 07:02 PM
h2osprey
Quote:

Originally Posted by Jose27
Is this supposed to be in the complex plane? because otherwise this doesn't make any sense (your $\displaystyle x \rightarrow 0^+$ puzzles me though) since $\displaystyle (\ln x )^x$ is not even defined (in $\displaystyle \mathbb{R}$ ) in $\displaystyle (0,1)$

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.
• Oct 9th 2009, 07:08 PM
Jose27
Quote:

Originally Posted by h2osprey
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 $\displaystyle y^x$ is only defined for positive $\displaystyle y$, because $\displaystyle y^x=e^{x\ln y}$ and $\displaystyle \ln x >0$ iff $\displaystyle x \in (1, \infty)$ therefore $\displaystyle (\ln x )^x$ is only defined in $\displaystyle (1, \infty)$
• Oct 9th 2009, 07:09 PM
h2osprey
Quote:

Originally Posted by Jose27
The number $\displaystyle y^x$ is only defined for positive $\displaystyle y$, because $\displaystyle y^x=e^{x\ln y}$ and $\displaystyle \ln x >0$ iff $\displaystyle x \in (1, \infty)$ therefore $\displaystyle (\ln x )^x$ is only defined in $\displaystyle (1, \infty)$

You made a mistake -- it's ln y, not ln x.
• Oct 9th 2009, 07:15 PM
Jose27
Quote:

Originally Posted by h2osprey
You made a mistake -- it's ln y, not ln x.

No, that logarithm refers to the one in your problem.
• Oct 9th 2009, 07:17 PM
h2osprey
Quote:

Originally Posted by Jose27
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.
• Oct 9th 2009, 07:19 PM
h2osprey
Quote:

Originally Posted by Jose27
The number $\displaystyle y^x$ is only defined for positive $\displaystyle y$, because $\displaystyle y^x=e^{x\ln y}$ and $\displaystyle \ln x >0$ iff $\displaystyle x \in (1, \infty)$ therefore $\displaystyle (\ln x )^x$ is only defined in $\displaystyle (1, \infty)$

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.
• Oct 9th 2009, 07:36 PM
Jose27
Quote:

Originally Posted by h2osprey
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 $\displaystyle \lim_{k \rightarrow \infty } \ \ln ( \frac{1}{2k+1} )^{ \frac{1}{2k+1} }$ not using the definition of $\displaystyle x^x for x \in \mathbb{R} ^+$ but the fact that negative numbers have odd roots. Right?
• Oct 9th 2009, 07:40 PM
mr fantastic
Quote:

Originally Posted by h2osprey
Find the lim as x tends to 0+ of (ln x)^x.

$\displaystyle (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}$. So you should first consider $\displaystyle \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ which has the indeterminant form $\displaystyle \frac{\infty}{\infty}$. Using l'Hopitals rule is an obvious thing to do.
• Oct 9th 2009, 07:56 PM
Jose27
$\displaystyle \lim_{k \rightarrow \infty } \ \ln ( \frac{1}{2k+1} )^{ \frac{1}{2k+1} }$$\displaystyle = \lim_{k \rightarrow \infty } \ (-\ln ( 2k+1 ))^{ \frac{1}{2k+1} }$$\displaystyle = -\lim_{k \rightarrow \infty } \ (\ln ( 2k+1 ))^{ \frac{1}{2k+1} }$$\displaystyle = -\lim_{x \rightarrow \infty } \ (\ln ( x ))^{ \frac{1}{x} }$$\displaystyle = -\lim_{x \rightarrow \infty} \ e^{ \frac{1}{x} \ln ( \ln x)}$$\displaystyle = -e^{ \lim_{x \rightarrow \infty} \ \frac{\ln ( \ln x)}{x} }$. Now use L'Hopital.

Quote:

$\displaystyle (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}$. So you should first consider $\displaystyle \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ which has the indeterminant form $\displaystyle \frac{\infty}{\infty}$. 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
• Oct 9th 2009, 08:46 PM
NonCommAlg
i agree with Jose27: $\displaystyle \lim_{x\to0+}(\ln x)^x$ does not exist because there is no right neighbourhood of $\displaystyle x = 0$ which is a subset of the domain $\displaystyle (\ln x)^x.$
• Oct 9th 2009, 08:52 PM
mr fantastic
• Oct 9th 2009, 09:00 PM
Jose27
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)
• Oct 10th 2009, 12:44 AM
h2osprey
Quote:

Originally Posted by mr fantastic
$\displaystyle (\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}$. So you should first consider $\displaystyle \lim_{x \rightarrow 0^+} x \ln (\ln x) = \lim_{x \rightarrow 0^+}\frac{\ln (\ln x) }{\frac{1}{x}}$ which has the indeterminant form $\displaystyle \frac{\infty}{\infty}$. Using l'Hopitals rule is an obvious thing to do.

I'm confused, isn't $\displaystyle \lim_{x \rightarrow 0^+}\ln (\ln x)$ undefined?
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