Prove that $\lim\limits_{s\to0^+}s^z=\lim\limits_{s\to 0^+} e^{z\ln s}=0$ where $z\in\mathbb C$ and $Re(z)>0$ Prove that $\displaystyle\lim_{s\to0^+}s^z=\lim_{s\to 0^+} e^{z\ln s}=0$ where $z\in\mathbb C$ and $Re(z)>0$
Using the $\epsilon-\delta$ definition we have
Let $\epsilon>0.$
We have to find $\delta>0$ such that 
$0<s<\delta$ implies $|e^{z\ln s}|<\epsilon$
I have no idea what to do next, could anyone help please?
Maybe a hint? or maybe 2 hints?
Note I am not asking for the proof since I know this site doesn't work like that.

Hint: What does the image of $\ln s$ for $s\in(0,\delta)$ look like in the complex plane? How does that vary with $\delta$.
What does the image of $z\ln s$ look like, given that $z$ has positive real part? How does that vary with $\delta$?
What is then the maximal size of $e^{z\ln s}$, given $\delta$?
