Theorem (Sequential Criterion)
Let \(a \in \mathbb{R}\). Let \(I\) be an open interval containing \(a\). Let \(f\) be a real function defined everywhere on \(I\) except possibly at \(a\). The following two statements are equivalent:
  1. \(\lim\limits_{x \to a} f(x) = L\).
  2. For all sequences \((x_n)\) such that \(x_n \neq a\) and \(\lim\limits_{x \to \infty} x_n = a\), it follows that \(\lim\limits_{x \to a}f(x_n) \rightarrow L\).

Proof

\((1)\Rightarrow (2)\): Suppose that \(\lim\limits_{x \to \infty} f(x) = L\). We know that for every \(\epsilon > 0\), we can find a \(\delta > 0\) such that if

$$ \begin{align*} 0 < |x - a| < \delta, \quad \text{ then } \quad |f(x) - L| < \epsilon \end{align*} $$

But we’re also given that \(\lim\limits_{x \to \infty} x_n = a\). This means that we can find an \(N \in \mathbb{N}\) such when \(n \geq N\)

$$ \begin{align*} |x_n - a| < \delta \end{align*} $$

Since \(x_n \neq a\), then we can just apply the function limit defintion. Recall that since we have \(|x_n - a| < \delta\), then we must have

$$ \begin{align*} |f(x_n) - L| < \epsilon \quad \text{ for all $n \geq N$} \end{align*} $$

Therefore, \(\lim\limits_{x \to \infty} f(x_n) = L\).

Side Note: So instead of the input to \(f\) being all the points in the neighborhood \(a-\delta, a+\delta\), now it's all the sequence terms starting at \(x_{N+1},x_{N+2},...\)

\((2)\Rightarrow (1)\): Assume (2) is true. But now to show (1), suppose for the sake of contradiction that (1) is false. That is, there exists a \(\epsilon_0 > 0\) such that for all \(\delta > 0\) the following statement

$$ \begin{align*} (0 < |x - a| < \delta, \quad \text{ then } \quad |f(x) - L| < \epsilon_0) \end{align*} $$

doesn’t hold. Then, for each \(\delta = \frac{1}{n}\) where \(n \in \mathbb{N}\), there is some point \(x_n \in I\) such that it satisfies the following conditions

$$ \begin{align*} 0 < |x_n - a| < \delta, \quad \text{ and } \quad |f(x_n) - L| \geq \epsilon_0 \end{align*} $$

Now the first condition gives us

$$ \begin{align*} 0 < |x_n - a| < \frac{1}{n} \end{align*} $$

But since we know \(\frac{1}{n} \rightarrow 0\) as \(n \rightarrow \infty\). Then by the Squeeze Theorem \(x_n - a \rightarrow 0\) and so \(x_n \rightarrow a\). But recall that by (2), \(f(x_n) \rightarrow L\) as \(n \rightarrow \infty\). In particular, \(|f(x_n) - L| < \epsilon_0\) for large \(n\). But this is a contradiction to the second condition.

References