Limit of Function by Convergent Sequences

From ProofWiki

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

[edit] 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.

$\blacksquare$

[edit] 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:

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