lim_{x->0-} (x^x): Why does this limit not exist?

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elvis

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I have this limit:
[math]\lim_{x \to 0^-}x^x[/math]The website says it's not defined, why?
Thanks in advance.
 
elvis said:
I have this limit:
[math]\lim_{x \to 0^-}x^x[/math]The website says it's not defined, why?
Thanks in advance.
Click to expand...
What website says that, and what does it say about it? It's right; but I don't like explaining what an unidentified source says, with no context.

Have you tried thinking about the function [imath]x^x[/imath], and what its graph looks like for negative values of x?

Consider, for example, [imath]x=-0.02[/imath] and [imath]x=-0.03[/imath]. What does that tell you about the nature of this function?
 
Steven G said:
Have you tried solving this limit. If yes, what happened? Can we see your work?
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I tried, but I'm still learning. Here's what I did:
[math]\lim_{x \to 0^-}x^x = \lim_{x \to 0^-}e^{x\ln x} = e^{\lim_{x \to 0^-}x\ln x}[/math]The limit at the exponent is:
[math]\lim_{x \to 0^-}x\ln x = \lim_{x \to 0^-}\dfrac{\ln x}{\frac1x} = \lim_{x \to 0^-}\dfrac{\frac1x}{-\frac1{x^2}} = \lim_{x \to 0^-}-x = 0[/math]So the limit is [math]e^0 = 1[/math]
 
Dr.Peterson said:
What website says that, and what does it say about it? It's right; but I don't like explaining what an unidentified source says, with no context.

Have you tried thinking about the function [imath]x^x[/imath], and what its graph looks like for negative values of x?

Consider, for example, [imath]x=-0.02[/imath] and [imath]x=-0.03[/imath]. What does that tell you about the nature of this function?
Click to expand...
The website is Symbolab, it doesn't explain anything, I just entered the limit and it said it doesn't exist:1678744361103.png
 
elvis said:
I tried, but I'm still learning. Here's what I did:
[math]\lim_{x \to 0^-}x^x = \lim_{x \to 0^-}e^{x\ln x} = e^{\lim_{x \to 0^-}x\ln x}[/math]The limit at the exponent is:
[math]\lim_{x \to 0^-}x\ln x = \lim_{x \to 0^-}\dfrac{\ln x}{\frac1x} = \lim_{x \to 0^-}\dfrac{\frac1x}{-\frac1{x^2}} = \lim_{x \to 0^-}-x = 0[/math]So the limit is [math]e^0 = 1[/math]
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This works for [imath]x\rightarrow 0+[/imath], but [imath]\log x[/imath] for negative [imath]x[/imath] is not defined in the domain of real numbers. Have you tried answering questions in post #2 ?
 
blamocur said:
This works for [imath]x\rightarrow 0+[/imath], but [imath]\log x[/imath] for negative [imath]x[/imath] is not defined in the domain of real numbers. Have you tried answering questions in post #2 ?
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You're right, I didn't consider the fact that [imath]\ln x[/imath] wasn't defined.
 
elvis said:
I tried, but I'm still learning. Here's what I did:
[math]\lim_{x \to 0^-}x^x = \lim_{x \to 0^-}e^{x\color{red}\ln x} = e^{\lim_{x \to 0^-}x\color{red}\ln x}[/math]The limit at the exponent is:
[math]\lim_{x \to 0^-}x\ln x = \lim_{x \to 0^-}\dfrac{\ln x}{\frac1x} = \lim_{x \to 0^-}\dfrac{\frac1x}{-\frac1{x^2}} = \lim_{x \to 0^-}-x = 0[/math]So the limit is [math]e^0 = 1[/math]
Click to expand...
elvis said:
You're right, I didn't consider the fact that [imath]\ln x[/imath] wasn't defined.
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The fact that you can't take the log of x, when it is negative is not the whole story; there might be another way to approach the problem.

Have you done what I suggested?
Dr.Peterson said:
Have you tried thinking about the function [imath]x^x[/imath], and what its graph looks like for negative values of x?

Consider, for example, [imath]x=-0.02[/imath] and [imath]x=-0.03[/imath]. What does that tell you about the nature of this function?
Click to expand...
This is really important! You need to know what the function does for negative values of x before you try using it.
 
Dr.Peterson said:
The fact that you can't take the log of x, when it is negative is not the whole story; there might be another way to approach the problem.

Have you done what I suggested?

This is really important! You need to know what the function does for negative values of x before you try using it.
Click to expand...
I just realised that for [imath]x < 1[/imath], [imath]x^x[/imath] is a complex number. So I have the answer to the original question, thank you for clarifying.
 
elvis said:
I just realised that for [imath]x < 1[/imath], [imath]x^x[/imath] is a complex number. So I have the answer to the original question, thank you for clarifying.
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Do you mean [imath]x<0[/imath] ?
 
elvis said:
I just realised that for [imath]x < {\color{red}0}[/imath], [imath]x^x[/imath] is a complex number. So I have the answer to the original question, thank you for clarifying.
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It's actually a little more complicated than that; for one of the two examples I gave it's real, and for the other it is not. The latter is true of more points than the former!

I would simply say that it is not continuously defined as a real-valued function, so limits (to the left of zero) don't make sense, and as a complex-valued (and many-valued!) function, it is very poorly behaved, so it has no limit. For a picture of how bad a function it is, see here.
 
What would just the sign of (-pi)^(-pi) be?. If you don't have an answer for that, then how can (-pi)^(-pi) be a well defined number?
 
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