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In attempting to solve $\lim_{x\to0}x^x$, I tried two different approaches. One is to convert $x^x$ into a complex function and solve the limit in $\mathbb{C}$. The other is to take the limit of the points of the dense sets in $\mathbb{R}^{-}$ and $\mathbb{R}^{+}$.

According to this article, $\lim_{x\to0}{x}^{x}$ can be converted into $\lim_{x\to{0}}|x|^{x}(\cos((2n+1)\pi x)+i\sin((2n+1)\pi x)$ where $n\in\mathbb{N}$ are the branches of complex logarithm.

This leads to $$\lim_{x\to0}{|x|}^{x}\lim_{x\to0}\cos((2n+1)\pi x)+i\lim_{x\to0}|x|^{x}\lim_{x\to0}\sin((2n+1)\pi x)=1$$

So using complex analysis $\lim_{x\to0}{x^x}=1$

However, if we take the points on real axis, where x-values of the complex function of $|x|^{x}(\cos((2n+1)\pi x)+i\sin((2n+1)\pi x)=a+0i$ (see this graph), we have the following domain. $$\left\{x=\left.-\frac{m}{2k+1}\right|m,k\in\mathbb{N}\right\}\bigcup{\mathbb{R}^{+}}$$

Which is divided into

$$x^x=\begin{cases} x^x & x>0\\ |x|^x & x=\left\{ -{2m\over 2k+1}\ |\ m, k \in \Bbb N\right\}\\ -|x|^{x} & x=\left\{ -{2m+1\over 2k+1}\ |\ m, k \in \Bbb N\right\}\ \\ \text{undefined} & x=\left\{ -{2m+1\over 2k}\ |\ m, k \in \Bbb N\right\}\bigcup \left\{\mathbb{R}^{-}\backslash \mathbb{Q}^{-}\right\} \end{cases}$$

Since $\left.-\frac{2m+1}{2k+1}\right|m,k \in \mathbb{N}$ and $\left.-\frac{2m}{2k+1}\right|m,k \in \mathbb{N}$ are dense sets; they can approximate arbitrarily close to any $x\in{\mathbb{R}}^{-}$. Thus a limit can exist if the subsets converge to the same value.

Hence $\lim_{x\to0}x^x$ exists if

$$\lim_{\left\{x\in-\frac{2m+1}{2k+1}\right\}\to0^{-}}x^x=\lim_{\left\{x\in-\frac{2m}{2k+1}\right\}\to0^{-}}x^x=\lim_{x\to0^{+}}{x^x}$$

Which is the same as

$$\lim_{x\to0^{-}}-|x|^x=\lim_{x\to0^{-}}|x|^x=\lim_{x\to0^{+}}x^x$$

However this equality fails since $\lim_{x\to0^{-}}-|x|^x=-1$ and the other limit are equal to $1$.

So using real analysis, $\lim_{x\to0}x^x$ does not exist.

I believe that the limit should be the same by real or complex analysis but I am no expert in either feild.

Did I do both approaches correctly? Does my answer depend on which analysis I use?

Community
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Arbuja
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  • $\lim_{z\to 0}z^z=\lim_{z\to 0}e^{z\log(z)}=e^{|z|e^{i\arg(z)}(\log(|z|)+i\arg(z)+i2\ell \pi )}=1$ – Mark Viola Sep 04 '16 at 18:15
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    @Dr.MV I think that is wrong. First, if $;z\in\Bbb R;$ then it must be $;z>0;$ so that $;\log z;$ is defined, and thus we only have a one sided limit, not the limit. If $;z\in\Bbb C;$ things get even trickier, since have infinite possible branches for the complex logarithm, all of which take out with them the point zero, and thus the limit when $;z\to 0;$ properly cannot exist as we naturally cannot make $;z\to 0;$ along that branch. – DonAntonio Sep 04 '16 at 18:17
  • @donantonio I disagree. Regardless of the sheet, the limit is $1$. And if $z \in \mathbb{R}$, that does not exclude the negative real axis. – Mark Viola Sep 04 '16 at 18:20
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    @Dr.MV Perhaps someone can kick in and apport some insights, but I think that when you take the limit after some definite branch has already been chosen, the limit when $;z\to 0;$ cannot exist as for that the function must be defined in a neighborhood (a circle in the complex plane) around that point, and this doesn't happen with $;z=0;$, the complex logarithm and any branch you choose. In the real case it is even simpler: the limit cannot exist as the function must be defined in some open interval around the point of tendency. Now, you can take the one-sided limit . – DonAntonio Sep 04 '16 at 18:26
  • The limit must be taken along a path that remains in the domain of definition. But even if the path moves from 1 sheet to the next, the limit is still $1$. – Mark Viola Sep 04 '16 at 18:31
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    @Dr.MV I disagree with that as, per definition, the limit, in case it exists, must exist no matter how we approach the given point, and that is why in real anaylis it is required that the tendency point be an inner point of the function's domain. Otherwise, only one sided limits can exist. In two variable analysis (which is, topologically, the same as the complex plane), the same is required: the tendency point must an inner one of the function's domain. In this case, you can say that on any branch of the logarithm (i.e., resctricting ourselves to it) the limit exists and etc. – DonAntonio Sep 04 '16 at 18:34
  • @Dr.MV What about the existence of cluster points? In topology you can use dense sets to approximate $c\in{R}^{-}$ regardless if $c$ is in the dense sets. If the desnse set coverge, the limit exists. We can't ignore topology.... – Arbuja Sep 04 '16 at 20:26
  • @Arbuja First, $x^x\ne |x|^x e^{i(2n+1)\pi x}$. Rather, if we choose the principal branch of the logarithm, then on the negative real axis, we have $$x^x=e^{x,\log(x)}=e^{-|x|,\log(|x|+i\pi)}=|x|^{-x}e^{-i\pi |x|}$$ which clearly approaches $1$ as $|x|\to 0$. – Mark Viola Sep 04 '16 at 21:06
  • Ok, then what is my mistake with real analysis? – Arbuja Sep 04 '16 at 21:08
  • Please refer to earlier comments. – Mark Viola Sep 04 '16 at 21:11
  • @Dr.MV If you have time you can look at this researh paper https://drive.google.com/file/d/0B_YK_xX-Yw1BdDktTFhudGJmVTQ/view?usp=sharing. – Arbuja Sep 04 '16 at 21:41
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    Read the first paragraph of Section 7 on Page 204. Its claim is identical to mine. All threads of $x^x$ emanate from $(1+i0,0)$. – Mark Viola Sep 05 '16 at 01:48
  • How is it with $(x+1)^{x+1}=\sum\limits_{k=0}^\infty \binom{x+1}{k}x^k$ with $x\to -1$ ? In this case there is no doubt that the limit is $1$. – user90369 Sep 08 '16 at 18:56
  • @user90369 This is a double sided limit. Otherwise from the positive side the limit is one. – Arbuja Sep 08 '16 at 19:28
  • @Arbuja: O.k.,thanks. E.g. $x\to -1-0$ is not possible because $|x|<1$. – user90369 Sep 08 '16 at 19:38
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    I have read it but it confuses me. - It's similar to the discussion about $\frac{1}{z}$ with $z\in\mathbb{C}$ and $z\to 0$. The value is $\infty$ and the direction is arbitrarily in the complex plane. Or more complicated: What means $0^i$? - I think we have strictly distinguish between the value and the direction. $|x|^x$ is clear, if we set $0^0:=1$ and $0^{i\cdot 0}:=1$(!). With $x:=re^{i\alpha}$ it's left to discuss $e^{i\alpha x}=e^{i\alpha\Re(x)-\alpha \Im(x)}$. $,e^{i\alpha\Re(x)}$ and $e^{-\alpha \Im(x)}$ are becomming $1$ for $x\to \pm 0$. Therefore one gets $1$. – user90369 Sep 08 '16 at 20:52
  • Note: I forgot to multiplicate with $e^{i2\pi k}$, means to set $x:=re^{i\alpha+i2\pi k}$, $k\in\mathbb{Z}$. But with $e^{i2\pi kx}$ we get the same as with $\alpha$, if we set $\alpha:=2\pi k$. – user90369 Sep 08 '16 at 21:11
  • @user90369 This is the case for the complex plane. But in real analysis the limit is undefined. In most cases if the real limit is undefined and the complex limit is ouside the real plane but this not the case for $\lim_{x\to0}x^x$. Hopefully you understood how I used real analysis for this limit. – Arbuja Sep 08 '16 at 22:02
  • @user90369 Thanks for verifying that my technique for complex analysis is correct. What about for real analysis? – Arbuja Sep 08 '16 at 22:03
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    The real limit is not defined a priori. That's right, because $a^0=\frac{a}{a}=1$ has only sense for $a\ne 0$. But I think we can define $0^0=1$ if there is no contradiction. It often makes sense. --- I have added a proof below (I hope that I haven't made a mistake) where I have avoided to use $0^0$. – user90369 Sep 09 '16 at 13:11
  • @DonAntonio I disagree with that as, per definition, the limit, in case it exists, must exist no matter how we approach the given point, and that is why in real analysis it is required that the tendency point be an inner point of the function's domain. But you are still wrong, because built into this very definition is the requirement that the paths being considered be in the domain of $f$. In other words, if the limit is independent of the paths that exist in the domain of $f$, then it exists. Usually, we consider points in the a domain that is an open neighbourhood, but this need not be... – Angel Oct 21 '21 at 12:02
  • @user90369 You are wrong. Defining $0^0$ is not necessary for computing this limit at all. Also, you are wrong regarding $a^0$. The definition of $a^0$ is not $\frac{a}{a}$, and it never has been. The definition of $a^0$ is the product of the 0-tuple $()$ containing $a$. This product is 1. The equation $a^0 = \frac{a}{a}, a \neq 0$ is a consequence of the definition (a.k.a a theorem), it is not itself a definition, and it also requires that one defines $a^{-1}$ as $\frac{1}{a}$. – Angel Oct 21 '21 at 12:05
  • @Angel Nop, you're wrong. The requirement that the paths must be within the function's domain is not necessarily built in ...though it should be in any decent definition. That's all what matters here. If that requirement is given then there is nothing to argue about. – DonAntonio Oct 30 '21 at 19:09
  • @DonAntonio In every rigorously given definition in analysis and topology, the paths being within the function's domain is built into the definition. I have no idea why you are in denial about this, but this much is true. I am not going to attempt discussing this with you any further. – Angel Nov 01 '21 at 11:44
  • @Angel Wise decision. – DonAntonio Nov 01 '21 at 22:16

2 Answers2

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Before we can start computing $\lim_{x\to0}f(x)$ we have to make sure that $f$ is well defined in a suitable, maybe restricted, neighborhood of $0$. Now powers $a^b$ with arbitrary real (or complex) exponents are bona fide (i.e., without additional explanations) defined only when the base $a$ is a positive real number. In this sense $x^x$ is well defined for $x>0$, and one has $\lim_{x\to0}x^x=\lim_{x\to0}\exp(x\log x)=e^0=1$.

Now it is customary to extend the $\log$ function from the positive real axis ${\mathbb R}_{>0}$ to the slit complex plane ${\mathbb C}':={\mathbb C}\setminus{\mathbb R}_{\leq0}$ by defining the principal value $${\rm Log}\,z:=\log|z|+i\,{\rm Arg}\,z\ ,$$ whereby $-\pi<{\rm Arg}\,z<\pi$ is the principal value of the polar angle of $z\in{\mathbb C}'$. For $a\in{\mathbb C}'$ and arbitrary $b\in{\mathbb C}$ one then defines $$a^b:=\exp(b\,{\rm Log}\,a)\ .$$ This definition extends the definition of "general powers" $(a,b)\mapsto a^b$ from ${\mathbb R}_{>0}\times{\mathbb R}$ to ${\mathbb C}'\times{\mathbb C}$. In this way we can consider $$z^z:=\exp\bigl(z(\log|z|+i\,{\rm Arg}\,z)\bigr)\qquad(z\in{\mathbb C}')\ .\tag{1}$$ While $0\notin{\mathbb C}'$ the origin is certainly a limit point of ${\mathbb C}'$. It is therefore allowed to consider the $\lim_{z\to0}$ in $(1)$. From the well known limit $\lim_{x\to0+}x\log x=0$ it then easily follows that $$\lim_{z\to 0}\bigl(z(\log|z|+i\,{\rm Arg}\,z)\bigr)=0\ ,$$ so that $\lim_{z\to0}z^z=1$. But note that we have excluded the negative real axis completely from the picture. If $\gamma: \ t\mapsto z(t)$ $(0\leq t<\infty)$ is a suitable spiral then using a continuous argument along $\gamma$ the statement $\lim_{t\to\infty}z(t)^{z(t)}=1$ may fail.

Christian Blatter
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  • @Christian Blatter : You are using $0^0=1$. What is the legitimacy for that ? I know: It's useful - but that's all. In opposite to the complex plane, where you can circle the point, you are comming on the real axis from the positve side to an undefined point (there is no continuation to the negative side of the axis). – user90369 Sep 09 '16 at 15:42
  • @Christian Blatter : Yes, I had seen, that $z\in\mathbb{C}'$. Therefore I was astonished that you have used $\lim\limits_{z\to 0},$. – user90369 Sep 09 '16 at 15:57
  • @Christian Blatter : I don't see any problem what you are concluding from $\lim\limits_{x\to 0^+} x\ln x$, the problem is $\lim\limits_{x\to 0^+} x\ln x$ itself. If you would say that's not a limit, it's an infimum/ a supremum, I would understand that it is useful to set $0^0=1$. But a limit ? No, because this means that there is an exact solution. In the complex case you can circle the value, but on the real axis you are comming only from one side - I think, that's the problem. With an exact solution we could say that $0^0=\frac{0}{0}=1$ which is obiously not true. – user90369 Sep 09 '16 at 16:24
  • @Christian Blatter You never stated if my approach using real analysis or complex analysis is correct. – Arbuja Sep 10 '16 at 11:24
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    I didn't understand why you brought integers $m$ and $n$ into the game. – Christian Blatter Sep 10 '16 at 11:58
  • The integers $n$ was to denote all the branches of the complex logarithm. The $m$ and $k$ integers represent all the point defined in $x^x$ when $x<0$. These dense subsets allow for the existence of limits on the negative real axis. – Arbuja Sep 11 '16 at 00:43
  • I've made edits – Arbuja Sep 11 '16 at 00:53
  • @user90369 Stop spreading misinformation. Christian never used $0^0 = 1$, and $0^0 = \frac{0}{0}$ is not true. – Angel Oct 21 '21 at 12:07
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In the complex plane $z^z$ is uniquely defined when $z=n\in {\Bbb Z}^*$ (i.e. a non-zero integer) and only for those. First note that any two definitions of $w'$ and $w$ of $z^z$ (for $z\neq 0$) must be related by the existence of some $k\in {\Bbb Z}$ for which: $$ w' = \exp \; \left(z \; (\log (z) + 2\pi i k)\;\right) = w e^{2\pi i k z}$$ so $kz\in {\Bbb Z}$ for all $k$ which means $z\in {\Bbb Z}$ (and zero was excluded). For any non-integer value of $z$ the mere definition of $z^z$ relies on your choice of logarithmic branch.

The discussion here is whether there is a unique way of defining it also when $n=0$. The answer is no in most generality. If, however, we approach zero radially then the limit is $1$. So the choice of $1$ is favored in some sense...

More precisely, consider a continuous curve $z: t\in (0,1] \mapsto z(t) \in {\Bbb C}^*$ with $z(1)=1$ and such that $\lim_{t\rightarrow 0^+} z(t)=0$. Let us also fix a logarithmic branch of $z$ so that $\log(z(1))=0$. This defines uniquely the logarithm along the curve: $$ \log z(t) = \log r(t) + i\,\phi(t) $$ where $\phi(1)=0$, $r(1)=1$, $r>0$ and $r(0^+)=0$. Then $$ z(t)\log z(t) = r(t) \left(\cos \phi(t) + i \sin \phi(t) \right) \left( \log r(t) + i\,\phi(t)\right) $$ Now $r(t)\log r(t)$ goes to zero as $r(t)\rightarrow 0^+$ so the question is reduced to the study of possible accumulation points of $$ r(t) \phi(t) \left( i \cos \phi(t) - \sin \phi(t) \right) $$ If you consider a ray $z(t)$ that approaches zero (asymptotically) radially then this corresponds to $\phi(t)$ having a limit as $t\rightarrow 0^+$ and then the above expression goes to zero. Thus for any radial limit, or more generally, when $r(t)\phi(t)$ goes to zero, we have $$ \lim_{t \rightarrow 0^+} z(t)^{z(t)} = \exp(0) = 1 $$ You may, on the other hand, choose $\phi(t)$ so that $r(t)\phi(t)$ converges to any real value (or diverges) as $t\rightarrow 0$. You may even (but this is somewhat more lengthy to describe) choose a continuous path $z(t)$ going to zero and such that $z(t)^{z(t)}$, $0<t\leq 1$ is dense in the complex plane!

H. H. Rugh
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  • This is true for complex analysis and it seems to check with real analysis but did I do the both ways correctly. Did I use real and complex analysis correctly as in my post? – Arbuja Sep 14 '16 at 14:08
  • I think your $k$ and $n$ are the same? But apart from that your analysis seems ok. It corresponds in my picture to let $\phi(t)r(t)=2m$ (or $2m+1$). But it means that you pick branches of log that winds around rapidly as $r\rightarrow 0$. As I mention, and you also indicate, you may obtain any limit doing that (e.g. $-1$). On the other hand, if you pick branches for which $\phi(t)$ stays bounded then the limit is always 1. – H. H. Rugh Sep 14 '16 at 14:50
  • No $n$ and $k$ are different integers. More specifically $2n+1=s(2k+1)$ where $s\in \mathbb{N}$. Since $s=\frac{2n+1}{2k+1}$, $s$ is odd. If $s$ is $2s+1$ then we substitute $2n+1$ with $(2s+1)(2k+1)$. You'll get all the defined reals. – Arbuja Sep 14 '16 at 18:55
  • Thank you for your answer your the only who bothered to check my post. – Arbuja Sep 14 '16 at 18:58
  • No problem. I think you get essentially the same results if you simply set $k=n$ (i.e. $s=1$). Anyway, in your post the relationship between $k$ and $n$ is not illucidated. – H. H. Rugh Sep 14 '16 at 19:10