The limit of $\frac{1}{z}$ in complex number

Question

How to find $\lim_{z\to z_0}\frac{1}{z}$?

I want to make a $\delta-\epsilon$ argument to prove $\lim_{z\to z_0}\frac{1}{z}=\frac{1}{z_0}$.

Let $\epsilon>0$. There is a $\delta = \frac{1}{\epsilon} > 0$ such that when $z \in \mathbb{C}\backslash \{0\}$ and $0<|z-z_0|<\delta$, we have $|f(z)-\frac{1}{z_0}|=|\frac{1}{z}-\frac{1}{z_0}|=|\frac{z_0-z}{z\cdot z_0}|=\frac{|z-z_0|}{|z \cdot z_0|}$

I have no idea on how to continue this proof.

Answer 1

You made a good start. Now, at first restrict $\delta$ to being less than $|z_0|/2$.

Then $|z| \ge |z_0|/2$. Use this to get rid of the $z$ in the denominator. Then your expression is of the form $C_{z_0}|z-z_0|$. Then chose $\delta$ to make this less than $\epsilon$.

Of course this assumes $z_0\neq 0$.