I am trying to prove $$\lim_{z \rightarrow 1} \frac{2z + 1}{z + 2}=1.$$
From the definition, given $\epsilon > 0$, I need to determine $\delta > 0$ such that $0< |z-1| <\delta$ implies $| \frac{2z + 1}{z + 2} - 1|<\epsilon$. Doing a little bit of algebra I need to prove. $$\delta<\epsilon |z+2|$$
Any Hints!
Use the fact that $\displaystyle \left| \frac{2z+1}{z+2} - 1 \right| = \left| \frac{z-1}{z+2} \right|.$
If $|z - 1| < 1$ then $|z+2| = |3 + (z-1)| \ge 3 - |z-1| \ge 2$, and if, in addition, $|z-1| < 2\epsilon$ then $$\left| \frac{z-1}{z+2} \right| < \frac{2\epsilon}{2} = \epsilon.$$
This means that if simultaneously $|z-1| < 1$ and $|z-1| < 2\epsilon$, then $\displaystyle \left| \frac{2z+1}{z+2} - 1 \right| < \epsilon.$
You can take $\delta = \min\{1,2\epsilon\}$.