Limit of $x^x$ as $x$ tends to $0$ I am trying to solve the following limit:
$$\lim \limits_{x\to0} x^x$$
The only thing that comes to mind is to write $x^x$ as $e^{x\ln{x}}$ and getting the right sided limit would be easy but I don't see how I could get the left sided one seeing that the $\ln$ is not defined for negative numbers.
Is there something I am missing or is there another way to go about it?
P.S.:I don't know anything about derivatives so please keep it to the limits.

Claim: $\lim_{z \to 0} z^z = 1$, no matter which branch of the logarithm is used to define $z^z$.
More rigorously: let $U \subset \Bbb{C}$ be any simply connected open set such that $0 \in \overline{U} \setminus U$. Since $U$ is simply connected, there exists a branch of the natural logarithm $\ln : U \to \Bbb{C}$ defined on $U$. This branch corresponds uniquely to a branch of $f(z) = z^z$ on $U$ given by defining $z^z := e^{z \ln z}$. No matter what branch is selected, we will have $$\lim_{z \to 0} z \ln z = 0,$$ as we can see by switching to polar form. Every $z \in U$ may be written uniquely as $z = r e^{i \theta}$, where $r := r(z) = |z|$ is independent of the choice of branch of $\ln$ and $\theta := \theta(z) = \operatorname{Im}(\ln(z))$ is uniformly bounded on all of $U$. If we write $\ln(z) := \ln(r(z)) + i \theta(z)$, then as $z \to 0$, $r(z) \to 0$ and
\begin{align*}
\lim_{z \to 0} z \ln z &= \lim_{r \to 0^+} re^{i \theta} (\ln(r) + i \theta) \\
= \lim_{r \to 0^+} [(r \ln r) e^{i \theta}] &+ \lim_{r \to 0^+} (i r \theta  e^{i \theta}) \\
= 0 &+ 0 = 0, \
\end{align*}
since $\lim_{r \to 0^+} (r \ln r) = 0$ and the functions of $\theta$ are bounded everywhere on $U$. It follows that $$\lim_{z \to 0} f(z) = \lim_{z \to 0} e^{z \ln z} = e^{\lim_{z \to 0} z \ln z} = e^0 = 1.$$
