Limit of Function by Convergent Sequences

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

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

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

Then $\displaystyle \lim_{x \to c} f \left({x}\right) = l$ iff, for each sequence $\left \langle {x_n} \right \rangle$ of points of $S$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$, it is true that $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$.

Proof

 * Suppose that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$.

Let $\epsilon > 0$.

Then by the definition of the limit of a function, $\exists \delta > 0: d_2 \left({f \left({x}\right), l}\right) < \epsilon$ provided $0 < d_1 \left({x, c}\right) < \delta$.

Now suppose that $\left \langle {x_n} \right \rangle$ is a sequence of points of $S$ such that such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$.

Since $\delta > 0$, from the definition of the limit of a sequence, $\exists N: \forall n > N: d_1 \left({x_n, c}\right) < \delta$.

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

That means $0 < d_1 \left({x_n, c}\right) < \delta$ by the definition of a metric.

But that implies that $d_2 \left({f \left({x_n}\right), l}\right) < \epsilon$.

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

Thus $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$.


 * Now suppose that for each sequence $\left \langle {x_n} \right \rangle$ of points of $S$ such that $\displaystyle \forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$, it is true that $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$.

What we will try to do is assume that it is not true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, and try to find a contradiction.

So, if it is not true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, then:


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

where $\land$ denotes logical and.

For all $n \in \N_{>0}$, define:
 * $S_n = \left\{{x \in S: 0 < d \left({x, c}\right) < \dfrac 1 n \land d_2 \left({f \left({x}\right), l}\right) \ge \epsilon}\right\}$

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

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

Then $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$, but it is not true that $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$.

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

Corollary
Let $f: \R \to \R$ be a real function.

The above result holds for $f$ tending to a limit both from the right and from the left:


 * $\displaystyle \lim_{x \to b^-} f \left({x}\right) = l \iff \forall \left \langle {x_n} \right \rangle: \lim_{n \to \infty} f \left({x_n}\right) = l$
 * $\displaystyle \lim_{x \to a^+} f \left({x}\right) = l \iff \forall \left \langle {x_n} \right \rangle: \lim_{n \to \infty} f \left({x_n}\right) = l$

where $f$ is defined on the open interval $\left({a \,.\,.\, b}\right)$.