# Definition:Limit of Complex Function

## Definition

The definition for the limit of a complex function is exactly the same as that for the general metric space.

Let $A_1, A_2 \subseteq \C$ be subsets of the complex plane.

Let $c$ be a limit point of $A_1$.

Let $f: A_1 \to A_2$ be a complex function from $A_1$ to $A_2$ defined everywhere on $A_1$ except possibly at $c$.

Let $L \in A_2$.

Then $\map f z$ is said to tend to the limit $L$ as $z$ tends to $c$, and we write:

$\map f z \to L$ as $z \to c$

or

$\ds \lim_{z \mathop \to c} \map f z = L$

if the following equivalent conditions hold.

This is voiced:

the limit of $\map f z$ as $z$ tends to $c$.

### Epsilon-Delta Condition

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall z \in A_1: 0 < \cmod {z - c} < \delta \implies \cmod {\map f z - L} < \epsilon$

That is, for every real positive $\epsilon$ there exists a real positive $\delta$ such that every point in the domain of $f$ within $\delta$ of $c$ has an image within $\epsilon$ of some point $L$ in the codomain of $f$.

### Epsilon-Neighborhood Condition

$\forall \map {N_\epsilon} L: \exists \map {N_\delta} c \setminus \set c: \map f {\map {N_\delta} c \setminus \set c} \subseteq \map {N_\epsilon} L$

where:

$\map {N_\delta} c \setminus \set c$ is the deleted $\delta$-neighborhood of $c$ in $M_1$;
$\map {N_\epsilon} L$ is the $\epsilon$-neighborhood of $L$ in $M_2$.

That is, for every $\epsilon$-neighborhood of $L$ in $A_2$, there exists a deleted $\delta$-neighborhood of $c$ in $A_1$ whose image is a subset of that $\epsilon$-neighborhood.

## Equivalence of Definitions

These definitions are seen to be equivalent by the definition of the $\epsilon$-neighborhood.

## Notes

1. $c$ does not need to be a point in $A_1$. Therefore $\map f c$ need not be defined. And even if $c \in A_1$, in may be that $\map f c \ne L$.
2. It is essential that $c$ be a limit point of $A_1$. Otherwise there would exist $\delta > 0$ such that $\set {z: 0 < \cmod {z - c} < \delta}$ contains no points of $A_1$. In this case the first condition would be vacuously true for any $L \in A_2$, which would not do.