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Calculus Level 2

$\large \lim_{x \to 0 } x^x = \ ?$

Right and left limits do not agree 1 Left or right limit do not exist 0

3 solutions

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Lucas Tell Marchi
Feb 3, 2015

First of all, note that

$x^{x} = e^{\ln(x^{x})} = e^{x \ln x}$

Also note that using L'Hospital

$\lim_{x\rightarrow 0} x \ln x = \lim_{x\rightarrow 0} \left ( -x^{2} \frac{1}{x} \right ) = - \lim_{x\rightarrow 0} x = 0$

And, therefore, as $\exp$ is a continuous function

$\lim_{x\rightarrow 0} x^{x} = \lim_{x\rightarrow 0} e^{x \ln x} = e^{\lim_{x\rightarrow 0} x \ln x} = e^{0} = 1$

This looks convincing, but my concern is that, since this is a two-sided limit, when we make use of the function $\ln(x)$ we will be working with a complex function when $x \lt 0$ , in which case we have to be careful about the use of L'Hospital's rule. I'm not saying that this is wrong, but to be rigorous we need to be mindful of the issues specific to complex functions.

@Janardhanan Sivaramakrishnan has posted a note that discusses this interesting topic in greater depth.

Brian Charlesworth - 6 years, 4 months ago

I completely agree. :) For this you may observe that $\ln(-x) = i \pi + \ln(x)$ , and then the derivative is the same, since $i \pi$ is a constant. I have thought about it just now, perhaps I'm wrong, but this would definitely be the way I'd try to approach this problem.

Lucas Tell Marchi - 6 years, 4 months ago

Yes, I think that is the best approach. I don't think there would be a problem with the use of L'Hospital's either, once it is confirmed that $\ln(x)$ satisfies the complex version of a differentiable function. Either way, it does appear that there is a strong case to conclude that the answer is $1$ .

Brian Charlesworth - 6 years, 4 months ago

@Brian Charlesworth – This article reminded me of my mathematical physics teacher, haha. Interesting how it is hard to find material about complex L'Hospital. Perhaps someone should post a note about it.

Lucas Tell Marchi - 6 years, 4 months ago

Janardhanan Sivaramakrishnan
Feb 4, 2015

Please have a look at this discussion

Melissa Quail
Feb 2, 2015

Anything to the power of 0 is 1 so substituting in 0 as x in the limit we find that the limit is equal to $0^0$ which is $\boxed{1}$ .

Moderator note:

This solution is wrong. $0^0 \ne 1$ .

Have to be careful here. The reason that for any nonzero number $a$ that $a^0 = 1$ is because $a^0$ can be written as $\frac{a^n}{a^n} = 1$ . If $a=0$ then that fraction is undefined. I actually don't know what the consensus is among mathematicians. I know some calculus textbooks refer to $0^0$ as an "indeterminate form."

Ryan Tamburrino - 6 years, 4 months ago

True. How would you approach it?

Melissa Quail - 6 years, 4 months ago

In a pinch, I'd probably answer this problem with $1$ just by looking at $x^x$ graphically.

Ryan Tamburrino - 6 years, 4 months ago

@Ryan Tamburrino – You can do that for the limit from the positive direction, but unfortunately not for the limit in the negative direction because of tricky technical things like complex exponentiation.

This problem is way underrated.

Daniel Liu - 6 years, 4 months ago

@Daniel Liu – I thought about that. Hm. Is there a consensus on this topic among mathematicians, or is it kind of up in the air?

Ryan Tamburrino - 6 years, 4 months ago

@Ryan Tamburrino – For this problem, there is a consensus. The answer is $1$ .

I quote from Calvin:

the left hand limit exists.

How do we conclude that?

somewhat painfully, because we're dealing with complex exponentiation you can show that the limit of the absolute value is 1 and that the limit of the argument theta is 0 (IE tan^-1 y/x )

Daniel Liu - 6 years, 4 months ago

@Daniel Liu – And now we get back to tetrations again. According to Wikipedia, (not the most reliable of sources, but whatever), $\lim_{x\rightarrow0} {}^{n}x$ equals $1$ when $n$ is even and $0$ when $n$ is odd. But as discussed before, the infinite tetration only exists for $e^{-\frac{1}{e}} \le x \le e^{\frac{1}{e}}$ , so there would be no solution for $x = 0$ .

P.S.., How does one do the tetration symbol in LaTeX? I had to put a dot before the exponent to make it work, but that makes it look like I've done something to the $0$ in my limit.

Brian Charlesworth - 6 years, 4 months ago

@Brian Charlesworth – Hmm... Have you tried just putting nothing before it? $^n x$

^n x

Or if you want to be really safe about it, put empty brackets before it: ${}^n x$

{}^n x

EDIT: I see why the first variety didn't work... it goes back to thinking that the power is meant for the previous element, and moves back there. In that case, use the second variant. I've fixed it for you on your post for you to see.

Daniel Liu - 6 years, 4 months ago

@Daniel Liu – Thanks! Yeah, I tried the first variant with unfortunate results, but your second variant does the trick. I'm assuming that you are a moderator, then, seeing that you can edit my comments. "With great powers comes great responsibility." :)

Brian Charlesworth - 6 years, 4 months ago

@Ryan Tamburrino – According to Michael Mendrin, as he states in the dispute section, there is no consensus. At the very least, this question is a good starting point for a discussion on complex analysis.

Brian Charlesworth - 6 years, 4 months ago

You cannot choose the path that you want to take, once a path has been given to you. In particular,

$\lim_{x \rightarrow 0} x^ 0 , \lim_{ x \rightarrow 0 } 0^x, \lim_{x \rightarrow 0 } x^x$

represent different things, as they are different paths of the $x^ y$ function.

Calvin Lin Staff - 6 years, 4 months ago

Yes, I agree with you, it's always a good idea to keep in mind that limiting value of the path is not necessarily the same as the end value. The former is path dependent. The end value is frequently indeterminate for the reason that there could be inconsistency among the limiting values of different paths taken, even though they could all end up at the same point.

The question of what's the value of $0^0$ is a separate matter, and while many mathematicians propose that it should be $1$ , it's indeterminate.

Michael Mendrin - 6 years, 4 months ago

With limits involving two independent variables, I've noticed that WolframAlpha is horribly unreliable, (it's wrong about half the time). It makes me wonder if this is because it is difficult to automate an analysis of path (in)dependency, or whether it just wasn't a priority of the WA programmers to figure it out. It's excellent when one real variable is involved, but I'm wary of its' results when it comes to complex numbers. With that caveat, it does return an answer of $1$ to the posted question.

Brian Charlesworth - 6 years, 4 months ago

x^0=1 and 0^x=0, So 0^0=0 or 1...

Lee Isaac - 6 years, 4 months ago

As mentioned, "You cannot choose the path that you want to take, once a path has been given to you. In particular,

$\lim_{x \rightarrow 0} x^ 0 , \lim_{ x \rightarrow 0 } 0^x, \lim_{x \rightarrow 0 } x^x$

represent different things, as they are different paths of the $x^ y$ function. "

Calvin Lin Staff - 6 years, 4 months ago

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