Statement and proof of the sequential characterization of limits

Question

So frequently on this site I will use the sequential characterizations of limits and continuity in my answers only to find the OP is unfamiliar with it. It is such a crucial notion that I am going to write the statement and proof here so that I (and others) can link it in answers.

Answer 1

Theorem (Sequential Characterization of Limits):

Let $I$ be an open interval, $a \in I$, and $f:I \to \mathbb{R}$, then:

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$$\lim_{x \to a}f(x) = L \text{ if and only if }\lim_{n \to \infty}f(x_n) = L \text{ for every sequence }x_n \in I - \{ a \} \text { satisfying } \\ x_n \to a \text{ as } n \to \infty$$

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Proof:

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$\Rightarrow:$

Suppose $\lim_{x \to a}f(x) = L$ and let $\{x_n\}$ be a sequence such that $x_n \to a$. Let $\varepsilon > 0$ be given. Choose $\delta > 0$ such that $$ 0 < \vert x-a\vert < \delta \implies \vert f(x) - L \vert < \varepsilon$$

Since $\delta > 0$ and $x_n \to a$ we can choose an $N \in \mathbb{N}$ such that $$ n \geq N \implies \vert x_n - a \vert < \delta$$

Combining both statements we have $$n \geq N \implies |f(x_n) - L| < \varepsilon$$

and hence $\lim_{n \to \infty}f(x_n) = L$

$\Leftarrow$:

By contradiction, suppose $x_n \to a$ implies $f(x_n) \to L$, but $ \ \lim_{x \to a}f(x) \neq L$, then $$\left(\exists \, \varepsilon_0 > 0\right)\left(\forall \delta > 0\right)\left(\exists x \in I \right)\left(0 < \vert x - a \vert < \delta \text{ but } \vert f(x) - L \vert \geq \varepsilon_0\right) $$

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Thus we can define a sequence $\{x_n\}$ as follows: For each $\delta = 1/n$, choose $x_n \in I - \{a\}$ such that: $$ 0 < \vert x_n - a \vert < 1/n \text{ and } \vert f(x_n) - L \vert \geq\varepsilon_0$$

By the squeeze theorem, $x_n \to a$ and thus by assumption $f(x_n) \to L$. However, by construction, $\vert f(x_n) - L \vert \geq \varepsilon_0$ for all $n$ and thus $\{f(x_n)\}$ cannot converge to $L$. We have arrived at a contradiction. This completes the proof.