Find the lim as x tends to 0+ of (ln x)^x.

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- Oct 9th 2009, 05:29 PMh2ospreyEvaluate a Limit
Find the lim as x tends to 0+ of (ln x)^x.

- Oct 9th 2009, 06:43 PMJose27
- Oct 9th 2009, 07:02 PMh2osprey
- Oct 9th 2009, 07:08 PMJose27
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 PMh2osprey
- Oct 9th 2009, 07:15 PMJose27
- Oct 9th 2009, 07:17 PMh2osprey
- Oct 9th 2009, 07:19 PMh2osprey
- Oct 9th 2009, 07:36 PMJose27
- Oct 9th 2009, 07:40 PMmr 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.

- Oct 9th 2009, 07:56 PMJose27
$\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.

- Oct 9th 2009, 08:46 PMNonCommAlg
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 PMmr fantastic
- Oct 9th 2009, 09:00 PMJose27
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 AMh2osprey