Limit $\lim \limits_{|z|\to 0} z^z$ and why $0^0 \neq 1$ I've always been puzzled by the value $0^0$. I remembered that one of my professor claimed that it was truely equal to $1$. However I think that most people would say, from an analytic point of view, that it is indeterminate which I agree with. Translating $0^0$ into $e^{0 \ln(0)}$ makes the problem appear explicitly. 
I also know the classic limit on $\mathbb{R}$, $\lim_{x\to 0^+} x^x = 1$ which could make us define by convention that $0^0$ is indeed $1$. However I once stumbled upon this Numberphile video in which (around 9:00) Matt Parker claims that even though the two sided limits $\lim_{x\to 0^{\pm}} x^x = 1$, when we consider the whole complex plan, this doesn't hold anymore and that's why we cannot say that $0^0 = 1$. Curious, I tried to explicitly prove that the limit doesn't exist. I don't have a good knowledge of complex analysis but here is what I tried :
Let $z \in \mathbb{C},\ z = r\ e^{i \theta}$ with $r \in \mathbb{R^+},\ \theta \in [0,2 \pi[$.
$$
\lim_{|z| \to 0} z^z = \lim_{r\to 0^+} (r\cdot e^{i\theta})^{r\cdot e^{i\theta}} =  \lim_{r\to 0^+} \overbrace{r^{re^{i\theta}}}^{(A)}\cdot \overbrace{e^{i\theta\cdot r e^{i\theta}}}^{(B)}
$$
We can cut the limit of product assuming both limits exist :
$$
(A) = \lim_{r\to 0^+} e^{\ln(r) r e^{i\theta}} = e^{\lim_{r \to 0^+} \ln(r) r e^{i\theta}} = e^0  = 1 \\
(B) = e^{\lim_{r\to 0^+} i\theta\cdot r e^{i\theta}} = e^0 = 1
$$
Therefore $$
\lim_{|z| \to 0} z^z = 1
$$
independently from $\theta$ which proves that the limit is $1$ everywhere in the complex plan. Is my reasoning wrong ? Is the video wrong ? Thank you !

Expanding a bit my question, since it seems that my reasoning is indeed wrong, could you also provide me an example of 2 paths in the complex plan which lead different limit values ?
 A: I have good and bad news for you. The good news: in the most applied scenarios, $z^z$ indeed converges to $1$ as $z$ converges $0$. The bad news: it is not so easy!
In your question you demonstrated, that approaching the zero on straight lines gives you $1$. But you should not forget the context in which you made this observation. At first, for complex numbers even the expression $z^w$ is not as easy as it seems. You have to use the defintion
$$z^w=\exp( w\log(z)).$$
This involves using the complex logarithm which cannot be defined on the whole complex plane $\mathbb C$ at the same time (if we are interested in $\log(z)$ beeing analytic). Instead, one cuts out a curve that connects $0$ and $\infty$. The remaining set $\mathbb C^*$ has an analytic definition of $\log(z)$, unique up to a constant term $2\pi k i,k\in\mathbb Z$. So choosing such a cut plane and such a term will give you also a unique definition of $\log(z)$ and $z^w$ or $-$ important in our case $-$ of $z^z$. Finally, for such a unique definition there is a unique continuous function $\phi:\mathbb C^*\rightarrow \mathbb R$ with
$$
\log(z)=\log|z|+i\phi(z).
$$
Note that studying the behavior of $z^z$ in $z=0$ is essentially the same as studying $z\log(z)$ ($="\!\log(z^z)\!"$) in $z=0$. The latter one converges for $z\rightarrow0$, if and only if $z^z$ converges and the limit $z^*$ of $z\log(z)$ gives a limit $\exp(z^*)$ for $z^z$. So the question is: $\lim_{z\rightarrow 0}{z\log(z)}=0$?
The answer turns out to be dependent only on the cut we made in order to define $\log(z)$. You, most probably, chose the principal branch of the complex logarithm, which is defined on $\mathbb C$ minus the non-positive real line. In this case $z\log(z)\rightarrow0$ (see below). But you have to exclude your path $r \exp(i\phi)$ with $\phi=\pi$ as this path is located directly inside the cut of $\mathbb C^*$. 
Lets look at approaching the zero in general with a curve $z_t$. We write $z_t=r_t\exp(i\phi(z_t))$, so $r_t\rightarrow 0$ with $t\rightarrow \infty$. We then have
\begin{align}
z_t\log(z_t) &= r_t\exp(i\phi(z_t))\cdot(\log r_t+i \phi(z_t))\\
&=\underbrace{r_t \log r_t}_{\rightarrow \,0} \cdot \exp(i\phi(z_t))+ r_t \phi(z_t) \cdot\exp(i\phi(z_t)).
\end{align}
So the behavior when approaching zero only depends on the second term (the first one vanishes; the behavior of $z\log(z)$ on the positive real line is known). In the limit, $r_t \phi(z_t)$ determines the absolute value of $z^*$ and $\phi(z_t)$ determines the argument of $z^*$. The following cases are possible:


*

*$z_t$ more or less directly approaches zero without winding itself around the origin, i.e. $\phi(z_t)$ is bounded. Then $r_t \phi(z_t)\rightarrow 0$, independent of the convergence of $\phi(z_t)$. This gives the usual result $z\log(z)\rightarrow 0$.

*$z_t$ slowly spirals around the origin, i.e. $\phi(z_t)$ is unbounded but in $\mathrm o(r_t^{-1})$. So still $r_t\phi_t\rightarrow 0$ and we have the usual result.

*$z_t$ spirals around the origin faster, i.e. $\phi(z_t)$ is unbounded and in $\Theta(r_t^{-1})$. But now, even if $r_t\phi_t$ convergec, the limit $z_t\log(z_t)$ does not exist as the argument does not converge. Also, $z_t\log(z_t)$ is not divergent, but oscillates.

*$z_t$ spirals around the origin even faster, i.e. $\phi(z_t)$ is unbounded and in $\Omega(r_t^{-1})$. Now, $r_t\phi_t\rightarrow\pm\infty$. $z_t\log(z_t)$ diverges to complex infinity (converges to the northpole of the Riemann sphere).


As you can see, none of these case gives convergence to a limit other than $0$. Finally, note that the number of turns (and the pace of turning) of $z_t$ around the origin is determined by the cut of $\mathbb C^*$. If the cut only turns around the origin a finite number of times, so will $z_t$. 

So the reason why you could only find $z\log(z)\rightarrow 0$ was, that you used the principal branch of the complex logarithm, where the cut is a straight line from the origin to the left. But there are indeed cuts giving you either no convergence, or even divergence.

Note that even if there is no curve $z_t\rightarrow 0$ that will give you any other suggestion than $z \log (z)=0$, there are indeed sequences $z_n\rightarrow 0$ that will give you any result you want, say $z_n\log(z_n)\rightarrow z^*=r^*\exp(i\phi^*)$. Choose $z_n$ and an appropriate cut so that
\begin{align}
\phi(z_n)&=2\pi n+\phi^*,\\
r_n&=\frac{r^*}{2\pi n}.
\end{align}
Then $r_n\phi(z_n)\rightarrow r^*$ and the arguments are at all times equivalent to $\phi^*$.
A: The real problem with defining $0^0=1$ is not the case $z^z$.  
We say $0^0$ is indeterminate because there are sequences $a_n, b_n$ with $$\lim a_n = 0,\quad\text{and}\quad\lim b_n = 0$$ but not $$\lim a_n^{b_n}=1.$$  These cases do not have $a_n = b_n$.  They have $a_n$ much smaller than $b_n$
This is similar to not defining
$$
\frac{0}{0}=1
$$
even though
$$
\lim_{z\to 0} \frac{z}{z} = 1
$$
