[4.2.5] Divergence Criterion for Functional Limits:
Let \(f: A \rightarrow \mathbb{R}\) and let \(c\) be a limit point of \(A\). If there exists two sequences \((x_n)\) and \((y_n)\) in \(A\) with \(x_n \neq c\) and \(y_n \neq c\) and $$ \begin{align*} \lim x_n = \lim y_n = c \quad \text{ but } \quad \lim f(x_n) \neq \lim f(y_n), \end{align*} $$ then we conclude that the functional limit \(\lim_{x \rightarrow x} f(x)\) does not exist.
Let \(f: A \rightarrow \mathbb{R}\) and let \(c\) be a limit point of \(A\). If there exists two sequences \((x_n)\) and \((y_n)\) in \(A\) with \(x_n \neq c\) and \(y_n \neq c\) and $$ \begin{align*} \lim x_n = \lim y_n = c \quad \text{ but } \quad \lim f(x_n) \neq \lim f(y_n), \end{align*} $$ then we conclude that the functional limit \(\lim_{x \rightarrow x} f(x)\) does not exist.
Definition of functional limits
Example (1)
Take the function
$$
\begin{align*}
f(x) =
\begin{cases}
0, & \text{if } x \leq 0 \\
1, & \text{if } x > 0
\end{cases}
\end{align*}
$$
We know that \(\lim\limits_{x \rightarrow 0}\) doesn’t exist. We can use the divergence corollary to show this by considering the sequences
$$
\begin{align*}
(x_n) &= \frac{1}{n} \quad \text{ and } \quad (y_n) = -\frac{1}{n}
\end{align*}
$$
Then, clearly as \(n \rightarrow \infty\), \((x_n) \rightarrow 0\) and \((y_n) \rightarrow 0\). However, observe that \(f(x_n) = 1\) while \(f(y_n) = 0\) for all \(n\). This means that \(\lim f(x_n) = 1\) while \(\lim f(y_n) = 0\). Therefore,
$$
\begin{align*}
\lim f(x_n) \neq \lim f(y_n)
\end{align*}
$$
Thus by 4.2.5, the limit doesn’t exist.
References