Limit of Function by Convergent Sequences

Theorem
Let $$f$$ be a real function defined on an open interval $$\left({a \, . \, . \, b}\right)$$, except possibly at the point $$c \in \left({a \, . \, . \, b}\right)$$.

Then $$\lim_{x \to c} f \left({x}\right) = l$$ iff, for each sequence $$\left \langle {x_n} \right \rangle$$ of points of $$\left({a \, . \, . \, b}\right)$$ such that $$\forall n \in \mathbb{N}^*: x_n \ne c$$ 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, $$\exists \delta > 0: \left|{f \left({x}\right) - l}\right| < \epsilon$$ provided $$0 < \left|{x - c}\right| < \delta$$.

Now suppose that $$\left \langle {x_n} \right \rangle$$ is a sequence of points of $$\left({a \, . \, . \, b}\right)$$ such that such that $$\forall n \in \mathbb{N}^*: x_n \ne c$$ and $$\lim_{n \to \infty} x_n = c$$.

Since $$\delta > 0$$, from the definition of the limit of a function, $$exists N: \forall n > N: \left|{x_n - c}\right| < \delta$$.

But $$\forall n \in \mathbb{N}^*: x_n \ne c$$.

That means $$0 < \left|{x_n - c}\right| < \delta$$.

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

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

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 $$\left({a \, . \, . \, b}\right)$$ such that $$\forall n \in \mathbb{N}^*: x_n \ne c$$ 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:

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

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

But then $$\left \langle {x_n} \right \rangle$$ is a sequence of points of $$\left({a \, . \, . \, b}\right)$$ such that $$\forall n \in \mathbb{N}^*: x_n \ne c$$ 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
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 $$\left({a \, . \, . \, b}\right)$$.