Prove that $\lim\limits_{s\to0^+}s^z=\lim\limits_{s\to 0^+} e^{z\ln s}=0$ where $z\in\mathbb C$ and $Re(z)>0$

Question

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.

Answer 1

$$e^{z\ln s}=\underbrace{e^{a\ln s}}_{(1)}\underbrace{\left(\cos(b\ln s)+i\sin(b\ln s)\right)}_{(2)}$$

Notice that in $(2)$ we have $\left|\cos(b\ln s)+i\sin(b\ln s)\right|=1$ So:$$0<s<\delta\implies|{e^{a\ln s}}{\left(\cos(b\ln s)+i\sin(b\ln s)\right)}|=|{e^{a\ln s}}|\cdot |{\left(\cos(b\ln s)+i\sin(b\ln s)\right)}|\\=|{e^{a\ln s}}|=e^{a\ln s}=s^a<\varepsilon$$

Now all it left to find is $\delta$ for $s^a$($a=\Re (z)>0$ is a constant)

Answer 2

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 $\delta$?

Answer 3

Use, for $z=x+iy$, $$ e^{z\ln s}=e^{x\ln s}e^{iy\ln s}= e^{x\ln s}(\cos(y\ln s)+i\sin(y\ln s)) $$ Since $x>0$ by assumption, we know that $$ \lim_{s\to0^+}e^{x\ln s}=0 $$ On the other hand, both $\cos(y\ln s)$ and $\sin(y\ln s)$ are bounded.