Is there a general solution for
$$\lim_{x\to0^+} x^z$$ for $z \in \mathbb{C}$. I assume we have to distinguish the cases $\mathrm{Re}[z]=0$, $\mathrm{Re}[z]>0$, $\mathrm{Re}[z]<0$, $\mathrm{Im}[z]=0$, $\mathrm{Im}[z]>0$, $\mathrm{Im}[z]<0$.
The case $\mathrm{Im}[z]=0$ is rather simple:
$$\lim_{x\to0^+} x^z = \begin{cases} \infty \quad \mathrm{Re}[z]<0 \\ 1 \,\,\,\quad \mathrm{Re}[z]=0 \\ 0 \,\,\,\quad \mathrm{Re}[z]>0 \end{cases}$$
But what are the solutions for the other cases?
You have to use the definition how a number with complex exponent is defined. I'm familiar with $$ x^z = e^{z\ln(x)}. $$ Now you can conlcude $$ x^z = e^{\Re(z)\ln(x)}e^{\Im(z)\ln(x)i} $$ and you can see what happens. If $\Im(z)=0$ then your angle is $0$ and you have just a real value. But if $\Im(z)\neq 0$ then $x^z$ rotates around $0$. If further $\Re(z)<0$ it goes to $0$. If $\Re(z)=0$ it runs on the unit sphere without limit and if $\Re(z)>0$ it blows up "to infinity".