# Math Help - Evaluate a Limit

1. ## Evaluate a Limit

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

2. 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 $x \rightarrow 0^+$ puzzles me though) since $(\ln x )^x$ is not even defined (in $\mathbb{R}$ ) in $(0,1)$

3. Originally Posted by Jose27
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.

4. 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 $y^x$ 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 $(1, \infty)$

5. Originally Posted by Jose27
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.

6. 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.

7. 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.

8. Originally Posted by Jose27
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.

9. 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 $\lim_{k \rightarrow \infty } \ \ln ( \frac{1}{2k+1} )^{ \frac{1}{2k+1} }$ not using the definition of $x^x for x \in \mathbb{R} ^+$ but the fact that negative numbers have odd roots. Right?

10. Originally Posted by h2osprey
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.

11. $\lim_{k \rightarrow \infty } \ \ln ( \frac{1}{2k+1} )^{ \frac{1}{2k+1} }$ $= \lim_{k \rightarrow \infty } \ (-\ln ( 2k+1 ))^{ \frac{1}{2k+1} }$ $= -\lim_{k \rightarrow \infty } \ (\ln ( 2k+1 ))^{ \frac{1}{2k+1} }$ $= -\lim_{x \rightarrow \infty } \ (\ln ( x ))^{ \frac{1}{x} }$ $= -\lim_{x \rightarrow \infty} \ e^{ \frac{1}{x} \ln ( \ln x)}$ $= -e^{ \lim_{x \rightarrow \infty} \ \frac{\ln ( \ln x)}{x} }$. 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

12. 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 $(\ln x)^x.$

13. 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)

14. Originally Posted by mr fantastic
$(\ln x)^x = e^{\ln (\ln x)^x} = e^{x \ln (\ln x)}$. 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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