Epsilon-delta exercise for limits of complex functions I'm trying to figure this one out.  This is an exercise from Churchill.  Show using epsilon-delta definition that:
$\lim_{z \to 1-i}x+i(2x+y)=1+i$
That is:
$|x+i(2x+y)-(1+i)|<\epsilon$ whenever $|z-(1-i)|<\delta$ for some $\delta$.
Can someone say if I did it right:
Choose $\delta$ so that $|x-1|<\frac{\epsilon}{6}$ and $|y+1|<\frac{\epsilon}{2}$.
Then 
$|x+i(2x+y)-(1+i)|=|x-1+i(2x+y-1)|\\
                  \leq |x-1|+|2x+y-1|\\
=|x-1|+|2(x-1)+y+1|\\
\leq |x-1|+2|x-1|+|y+1|\\
\leq 3|x-1|+|y+1|\\
\leq 3\frac{\epsilon}{6} + \frac{\epsilon}{2}=\epsilon$

Your reasoning looks correct, but you haven't prescribed a $\delta $ based on the given $\varepsilon$.
If suffices to give a formula for $\delta $ in terms of the specified $\varepsilon$. Since you want $\vert x-1 \vert < \frac{\varepsilon}{6}$ and $\vert y+1 \vert<\frac{\varepsilon}{2}$, take $\delta=\frac{\varepsilon}{6}$. Then, from your chain of inequalities, $$\vert (x-1)+i(2x+y-1)\vert\le 3\vert x-1\vert+\vert y+1\vert \\ <3\frac{\varepsilon}{6}+\frac{\varepsilon}{6}\\=\frac{2\varepsilon}{3}\\<\varepsilon$$
