So, I'm having trouble performing the $\varepsilon$ - $\delta$ for this proof.
I made the change from $z$ to $\frac{1}{z}$ so the limit becomes $\lim_{z \to 0} \frac{4}{(1-z)^2} = 4$.
Next, I applied the definition for the limit: $ \left| \frac{4}{(1-z)^2} - 4 \right| < \varepsilon$ when $\left| z \right| < \delta$. Then, after performing some math I get to $ \left| \frac{-z^2 +2z}{(1-z)^2}\right| < \frac{|z||z-2|}{|z-1|} < \varepsilon $.
Finally I choose $\delta<5$ that makes $ z-2 < 3$ and $ z-1 < 4$ and $ \delta $ is min{$5, \frac{4 \varepsilon}{3} $}.
I don't know if that is right or if there is a better way to do it.
Edit: This problem is from "Complex variables and applications" Brown & Churchill
When you get to the step of setting an upper bound on $\delta$ it needs to be chosen so that $\left| \frac {z-2}{z-1}\right|$ is bounded.
$\delta \le 4$ will not do the trick! We can't let the denominator equal zero.
Choose $\delta \le \frac 12$ (we could choose any value less than 1, but this is easy to work with)
In wich case $\left |\frac {z-2}{z-1}\right|< 5$
$\delta = \min (\frac \epsilon 5, \frac 12)$