real analysis - Prove: Given $x_0$ is a cluster point of a set $S$ and $f:S to mathbb{R}$ then $f$ can have at most one limit as $x to x_0$

Essentially, I need to prove that, given a point $x_0$ in $S$ where $S \subseteq \mathbb{R}$, as the $x$ value converges to $x_0$, $f(x)$ converges to only one $f(x_0)$. This is poking at the idea that for every input, $x$, there cannot be more than one output, $f(x)$.

This is something that most of us knew from algebra 1, but I need to prove this statement using the definition of cluster points, continuity, etc.

Cluster point: if $x_0$ is a cluster point, then $\forall \epsilon \gt 0$, $(x_0- \epsilon , x_0+ \epsilon ) \cap (S \setminus {x_0}) \neq \phi$

definition of continuity: $\forall \epsilon \gt 0$, $\exists \delta \gt 0$ such that $0 \lt |x-x_0| \lt \delta$, $x \in S$, implies $|f(x)-f(x_0)| \lt \epsilon$.

I'm not too sure how I would approach this!

Answer

First, the existence of limit at a point (say, $x_0$) does not depend on the value of the function at $x_0$. This is important because, without this idea, limit and continuity become more or less useless.

Now to show that $f$ has at most one limit as $x \to x_0$, we need to show that either the limit does not exist ("zero" limit), or the limit is unique if it exists ("one" limit).

In other words, we can assume that the limit exists, and let

$$\lim_{x \to x_0} f(x) = l_1$$ and $$\lim_{x \to x_0} f(x) = l_2$$ . Now we want to show that $l_1 = l_2$. Start from the definition.

Given $\epsilon > 0$, $\exists \delta_1 > 0$ such that $$|f(x)-l_1| < \frac{\epsilon}{2}, \forall x \in V_{\delta_1}(x_0)\cap{S}\setminus{\{x_0 \}}$$ Also, $\exists \delta_2>0$ such that $$|f(x)-l_2|<\frac{\epsilon}{2}, \forall x \in V_{\delta_2}(x_0)\cap{S}\setminus{\{x_0\}}$$

. Now take $\delta = min\{\delta_1, \delta_2\}$.

$$|l_1-l_2| = |l_1-f(x)+f(x)-l_2| \le |l_1-f(x)|+|f(x)-l_2| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon, \forall x \in V_\delta(x_0) \cap S\setminus {\{x_0\}}$$ . Since $x_0$ is a cluster point, $|l_1-l_2|$ is arbitrary small (but still non-negative) on a non-empty set. Can you conclude from here? (An extra hint: Suppose $\forall \epsilon > 0, 0\le a < \epsilon$, can you prove by contradiction that $a = 0$?)