Limit of Function by Convergent Sequences

Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $S \subseteq A_1$ be an open set of $M_1$.

Let $f: S \to A_2$ be a mapping defined on $S$, except possibly at the point $c \in S$.

Then $\ds \lim_{x \mathop \to c} \map f x = l$ :
 * for each sequence $\sequence {x_n}$ of points of $S$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$

it is true that:
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$

Real Number Line
On the real number line, this result becomes as follows:

Corollary
The above result holds for a real function tending to a limit both from the right and from the left:

Complex Plane
On the complex plane, this result becomes as follows:

Necessary Condition
Suppose that:
 * $\ds \lim_{x \mathop \to c} \map f x = l$

Let $\epsilon > 0$.

Then by the definition of the limit of a mapping:
 * $\exists \delta > 0: \map {d_2} {\map f x, l} < \epsilon$

provided $0 < \map {d_1} {x, c} < \delta$.

Now suppose that $\sequence {x_n}$ is a sequence of points of $S$ such that such that:
 * $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$

Since $\delta > 0$, from the definition of the limit of a sequence:
 * $\exists N: \forall n > N: \map {d_1} {x_n, c} < \delta$

But:
 * $\forall n \in \N_{>0}: x_n \ne c$

That means:
 * $0 < \map {d_1} {x_n, c} < \delta$

by the definition of a metric.

But that implies:
 * $\map {d_2} {\map f {x_n}, l} < \epsilon$

That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
 * $\forall n > N: \map {d_2} {\map f {x_n}, l} < \epsilon$

Thus:
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$

Sufficient Condition
Now suppose that for each sequence $\sequence {x_n}$ of points of $S$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$, it is true that:
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$

it is not true that:
 * $\ds \lim_{x \mathop \to c} \map f s = l$

So:
 * $\exists \epsilon > 0: \forall \delta > 0: \exists x \in S: 0 < \map {d_1} {x, c} < \delta \land \map {d_2} {\map f x, l} \ge \epsilon$

where $\land$ denotes logical and.

For all $n \in \N_{>0}$, define:
 * $S_n = \set {x \in S: 0 < \map {d_1} {x, c} < \dfrac 1 n \land \map {d_2} {\map f x, l} \ge \epsilon}$

By hypothesis, $S_n$ is non-empty for all $n \in \N_{>0}$.

Using the axiom of countable choice, there exists a sequence $\sequence {x_n}$ of elements of $S$ such that $x_n \in S_n$ for all $n \in \N_{>0}$.

Then:
 * $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$

but it is not true that:
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$

So there is our contradiction, and so the result follows.