Show that $\frac{1}{x}$ is not uniformly continuous on $(0, 1)$

I want to use the following criterion:

Let $f \colon A \to \mathbb R$ be a function. If there are sequences $x_n, y_n$ in $A$ such that $|x_n - y_n| \to 0$, and $|f(x_n) - f(y_n)| \to a \neq 0 $, as $n \to \infty$, then $f$ is not uniformly continuous on $A$.

Let $x_n=\frac{1}{2n}, y_n=\frac{1}{3n}$, then $$\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}\frac{1}{2n}=0=\lim\limits_{n\to\infty}\frac{1}{3n}=\lim\limits_{n\to\infty}y_n$$

hence $|x_n - y_n| \to 0$, as $n \to \infty$.

Now since $f(x_n) = \frac{1}{\frac{1}{2n}} = 2n$ and $f(y_n) = \frac{1}{\frac{1}{3n}} = 3n$

we have

$$\lim\limits_{n\to\infty}f(x_n)=\lim\limits_{n\to\infty}2n=+\infty=\lim\limits_{n\to\infty}3n=\lim\limits_{n\to\infty}f(y_n)$$

and thus $|f(x_n) - f(y_n)| \to \infty \neq 0$, which means $f$ is not uniformly continuous on $(0, 1)$.

I guess I'm essentially asking if the $a$ in the quoted criterion above can be $\pm \infty$, i.e. is $a \in \mathbb R \cup \{ - \infty, +\infty \}$.