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

Proof

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

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 $$\forall n \in \N^*: x_n \ne c$$ and $$\lim_{n \to \infty} x_n = c$$.

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 $$\forall n \in \N^*: x_n \ne c$$.

That means $$0 < d_1 \left({x_n, c}\right) < \delta$$.

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


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

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

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

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

In particular, if $$\delta = \frac 1 n$$, we can find an $$x_n$$ where $$0 < d_2 \left({x, c}\right) < \frac 1 n$$ such that $$d_2 \left({f \left({x_n}\right), l}\right) \ge \epsilon$$.

But then $$\left \langle {x_n} \right \rangle$$ is a sequence of points of $$S$$ such that $$\forall n \in \N^*: x_n \ne c$$ and $$\lim_{n \to \infty} x_n = c$$, but for which it is not true that $$\lim_{n \to \infty} f \left({x_n}\right) = l$$.

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:


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

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