Limit as $x \to 0^-$ of $x^x$ from the negative side using complex numbers We know that $\lim_\limits{x \to 0^+}x^x = 1$. We also know that the limit as $x \to 0^-$ does not exist, at least for $x \in \mathbb{R}$.
When considering complex numbers, for $z\in \mathbb{C}$, does the following limit converge?
$$\lim_\limits{z \to 0^-}z^z$$
My attempt: Separate $z$ into its real and imaginary components, making this a multivariable limit. Let $\text{Re}(z)=\sigma,$ and $\text{Im}(z) = t$. Thus, we have
$$\lim_\limits{(\sigma, t) \to 0^-}(\sigma + it)^{\sigma+it}$$
But I'm not sure where to go from here.
 A: Your attempt is nice but it would make it more complex. You don't have to split the complex number $z$ into real and imaginary parts. That just creates more paperwork.
You should rather use the power laws and polar form until you get the same starting form as $\lim_\limits{{z} \to {0^{+}}} z^{z}$ then you can find the limit quite analogously to $\lim_\limits{{z} \to {0^{+}}} z^{z}$. If you do that, you can see that they are the same: $\lim_\limits{{z} \to {0^{-}}} z^{z} = \lim_\limits{{z} \to {0^{+}}} z^{z}$
$$
\begin{align*}
\lim_\limits{{z} \to {0^{-}}} z^{z} &= \lim_\limits{{z} \to {0^{-}}} \exp\left( \ln(z) \right)^{z}\\
\lim_\limits{{z} \to {0^{-}}} z^{z} &= \lim_\limits{{z} \to {0^{-}}} \exp\left( \ln(z) \cdot z \right)\\
\lim_\limits{{z} \to {0^{-}}} z^{z} &= \exp\left( \lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z \right)\\
\lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z &= \lim_\limits{{z} \to {0^{-}}} \frac{\ln(z)}{\frac{1}{z}}\\
\lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z &= \lim_\limits{{z} \to {0^{-}}} \frac{\ln(|z| \cdot \exp(\arg(z) \cdot \mathrm{i}))}{\frac{1}{z}}\\
\lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z &= \lim_\limits{{z} \to {0^{-}}} \frac{\ln(|z|) + \ln(\exp(\arg(z) \cdot \mathrm{i}))}{\frac{1}{z}}\\
\lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z &= \lim_\limits{{z} \to {0^{-}}} \frac{\ln(|z|) + \arg(z) \cdot \mathrm{i}}{\frac{1}{z}}\\
\lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z &= \lim_\limits{{z} \to {0^{-}}} \frac{\ln(|z|)}{\frac{1}{z}} + \lim_\limits{{z} \to {0^{-}}} \frac{\arg(z) \cdot \mathrm{i}}{\frac{1}{z}}\\
\lim_\limits{{z} \to {0^{-}}} \frac{\ln(|z|)}{\frac{1}{z}} &= \lim_\limits{{z} \to {0^{-}}} \frac{\frac{\operatorname{d}}{\operatorname{d}z} \ln(|z|)}{\frac{\operatorname{d}}{\operatorname{d}z} \frac{1}{z}} = \lim_\limits{{z} \to {0^{-}}} \frac{\frac{1}{|z|}}{-\frac{1}{z^{2}}} = \lim_\limits{{z} \to {0^{-}}} -\frac{z^{2}}{|z|} = \lim_\limits{{z} \to {0^{-}}} -\frac{z^{2}}{\frac{z}{\operatorname{sgn}(z)}} = \lim_\limits{{z} \to {0^{-}}} -\operatorname{sgn}(z) \cdot z = -0 \cdot 0 = 0\\
\lim_\limits{{z} \to {0^{-}}} \frac{\arg(z) \cdot \mathrm{i}}{\frac{1}{z}} &= \lim_\limits{{z} \to {0^{-}}} z \cdot \arg(z) \cdot \mathrm{i} = 0 \cdot (\theta + 2 \cdot k \cdot \pi) \cdot \mathrm{i} = 0, \theta \in \mathbb{R}\\
\lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z &= \lim_\limits{{z} \to {0^{-}}} \frac{\ln(|z|)}{\frac{1}{z}} + \lim_\limits{{z} \to {0^{-}}} \frac{\arg(z) \cdot \mathrm{i}}{\frac{1}{z}} = 0 + 0 = 0\\
\lim_\limits{{z} \to {0^{-}}} z^{z} &= \exp\left( \lim_\limits{{z} \to {0^{-}}} \ln(z) \cdot z \right) = \exp\left( 0 \right) = 1\\
\end{align*}
$$
