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

[elvis]

I have this limit:
$$\lim_{x \to 0^-}x^x$$The website says it's not defined, why?
Thanks in advance.

 

[Dr.Peterson]

I have this limit:
$$\lim_{x \to 0^-}x^x$$The website says it's not defined, why?
Thanks in advance.

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 $x^x$, and what its graph looks like for negative values of x?

Consider, for example, $x=-0.02$ and $x=-0.03$. What does that tell you about the nature of this function?

 

[Steven G]

I have this limit:
$$\lim_{x \to 0^-}x^x$$The website says it's not defined, why?
Thanks in advance.

Have you tried solving this limit. If yes, what happened? Can we see your work?

 

[elvis]

Have you tried solving this limit. If yes, what happened? Can we see your work?

I tried, but I'm still learning. Here's what I did:
$$\lim_{x \to 0^-}x^x = \lim_{x \to 0^-}e^{x\ln x} = e^{\lim_{x \to 0^-}x\ln x}$$The limit at the exponent is:
$$\lim_{x \to 0^-}x\ln x = \lim_{x \to 0^-}\dfrac{\ln x}{\frac1x} = \lim_{x \to 0^-}\dfrac{\frac1x}{-\frac1{x^2}} = \lim_{x \to 0^-}-x = 0$$So the limit is $$e^0 = 1$$

 

[elvis]

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 $x^x$, and what its graph looks like for negative values of x?

Consider, for example, $x=-0.02$ and $x=-0.03$. What does that tell you about the nature of this function?

The website is Symbolab, it doesn't explain anything, I just entered the limit and it said it doesn't exist:

 

[blamocur]

I tried, but I'm still learning. Here's what I did:
$$\lim_{x \to 0^-}x^x = \lim_{x \to 0^-}e^{x\ln x} = e^{\lim_{x \to 0^-}x\ln x}$$The limit at the exponent is:
$$\lim_{x \to 0^-}x\ln x = \lim_{x \to 0^-}\dfrac{\ln x}{\frac1x} = \lim_{x \to 0^-}\dfrac{\frac1x}{-\frac1{x^2}} = \lim_{x \to 0^-}-x = 0$$So the limit is $$e^0 = 1$$

This works for $x\rightarrow 0+$, but $\log x$ for negative $x$ is not defined in the domain of real numbers. Have you tried answering questions in post #2 ?

 

[elvis]

This works for $x\rightarrow 0+$, but $\log x$ for negative $x$ is not defined in the domain of real numbers. Have you tried answering questions in post #2 ?

You're right, I didn't consider the fact that $\ln x$ wasn't defined.

 

[Dr.Peterson]

I tried, but I'm still learning. Here's what I did:
$$\lim_{x \to 0^-}x^x = \lim_{x \to 0^-}e^{x\color{red}\ln x} = e^{\lim_{x \to 0^-}x\color{red}\ln x}$$The limit at the exponent is:
$$\lim_{x \to 0^-}x\ln x = \lim_{x \to 0^-}\dfrac{\ln x}{\frac1x} = \lim_{x \to 0^-}\dfrac{\frac1x}{-\frac1{x^2}} = \lim_{x \to 0^-}-x = 0$$So the limit is $$e^0 = 1$$

You're right, I didn't consider the fact that $\ln x$ wasn't defined.

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?

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

Consider, for example, $x=-0.02$ and $x=-0.03$. What does that tell you about the nature of this function?

This is really important! You need to know what the function does for negative values of x before you try using it.

 

[elvis]

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.

I just realised that for $x < 1$, $x^x$ is a complex number. So I have the answer to the original question, thank you for clarifying.

 

[blamocur]

I just realised that for $x < 1$, $x^x$ is a complex number. So I have the answer to the original question, thank you for clarifying.

Do you mean $x<0$ ?

 

[Dr.Peterson]

I just realised that for $x < {\color{red}0}$, $x^x$ is a complex number. So I have the answer to the original question, thank you for clarifying.

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.

 

[Steven G]

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?

 

[elvis]

I just realised that for $x < 1$, $x^x$ is a complex number. So I have the answer to the original question, thank you for clarifying.

Do you mean $x<0$ ?

Yeah, I can't edit the post anymore.