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


Which is the same as


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?

  • 3
    $\begingroup$ @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. $\endgroup$
    – DonAntonio
    Commented Sep 4, 2016 at 18:17
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    $\begingroup$ @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 . $\endgroup$
    – DonAntonio
    Commented Sep 4, 2016 at 18:26
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    $\begingroup$ @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. $\endgroup$
    – DonAntonio
    Commented Sep 4, 2016 at 18:34
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    $\begingroup$ Read the first paragraph of Section 7 on Page 204. Its claim is identical to mine. All threads of $x^x$ emanate from $(1+i0,0)$. $\endgroup$
    – Mark Viola
    Commented Sep 5, 2016 at 1:48
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    $\begingroup$ 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 $1$. $\endgroup$
    – user90369
    Commented Sep 8, 2016 at 20:52

2 Answers 2


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.

  • $\begingroup$ @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). $\endgroup$
    – user90369
    Commented Sep 9, 2016 at 15:42
  • $\begingroup$ @Christian Blatter : Yes, I had seen, that $z\in\mathbb{C}'$. Therefore I was astonished that you have used $\lim\limits_{z\to 0}\,$. $\endgroup$
    – user90369
    Commented Sep 9, 2016 at 15:57
  • $\begingroup$ @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. $\endgroup$
    – user90369
    Commented Sep 9, 2016 at 16:24
  • $\begingroup$ @Christian Blatter You never stated if my approach using real analysis or complex analysis is correct. $\endgroup$
    – Arbuja
    Commented Sep 10, 2016 at 11:24
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    $\begingroup$ I didn't understand why you brought integers $m$ and $n$ into the game. $\endgroup$ Commented Sep 10, 2016 at 11:58

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!

  • $\begingroup$ 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? $\endgroup$
    – Arbuja
    Commented Sep 14, 2016 at 14:08
  • $\begingroup$ 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. $\endgroup$
    – H. H. Rugh
    Commented Sep 14, 2016 at 14:50
  • $\begingroup$ 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. $\endgroup$
    – Arbuja
    Commented Sep 14, 2016 at 18:55
  • $\begingroup$ Thank you for your answer your the only who bothered to check my post. $\endgroup$
    – Arbuja
    Commented Sep 14, 2016 at 18:58
  • $\begingroup$ 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. $\endgroup$
    – H. H. Rugh
    Commented Sep 14, 2016 at 19:10

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