Here what i already done:
\begin{align*} |az+b-(az_0+b)&+|az-az_0+b-b|\\&=|a(z-z_0)|\\&=|a||z-z_0|{\le}|a|*{\Delta}|z-z_0|
\end{align*}
If ${\Delta}=1$ and we know that $0{\lt}|z-z_0|{\lt}1$, then:
$||z|-|z_0||{\le}|z-z_0|<1 \\{\to} |z|{\lt}1+|z_o|$
Then:
$|z|<1+|z_0| {\to} |z+z_0|{\le}|z|+|z_0|{\le}1+2|z_0|$
So:
${\Delta}|a|*(1+2|z_0|)\\ = {\Delta}(|a|*(1+2|z_0|)={\epsilon}\\ {\to} {\Delta} = {\epsilon}/(|a|(|1+2|z_0|)$
Then:
$|az+b – (az_0 + b)| {\lt} |a|{\Delta}|z-z_0|\\ {\le} {\Delta}(|a|(1+2|(z_0)|\\{\lt}{\epsilon}/(|a|(1+2|z_0|)*(|a|(1+2|z_0|) = {\epsilon}$
That proves:
$\lim_{z\to z_0} (az + b) = az_0 + b$
Is that correct?
Best Answer
Seems okay in general (but your approach is more complex than it should), just some fixes.
First of all, you need to consider the case $a=0 \in \mathbb C$. Then, if that's the case, you have nothing to prove, as it's straightforward. Then you can go on by considering the case $a\neq 0 \in \mathbb C$.
But for a more fast and elegant approach, observe that
$$|az + b - (az_0 + b)| = |a||z-z_0| < \varepsilon \implies |z-z_0| < \frac{\varepsilon}{|a|+1}$$
considering $a=0$ and $a\neq 0$ at the same time, thus you're finished.