If $x_0$ is a cluster point of a set $S$ and $f:S \to \mathbb{R}$ then $f$ can have at most one limit as $x \to x_0$

Question

I'm not too sure how to approach this type of question.

Definition of cluster point of $S$: $\forall \epsilon \gt 0$, $(x_0- \epsilon , x_0+ \epsilon ) \cap (S \setminus {x_0}) \neq \phi $

A real number $L$ is called the limit of $f$ at $x_0$ is defined as $\forall \epsilon \gt 0$, $\exists \delta \gt 0$ such that $0 \lt |x-x_0| \lt \delta$, $x \in S$, implies $|f(x)-L| \lt \epsilon$.

We denote this by $\lim_{x\to x_0} f(x)=L$

Essentially, it wants me to prove that there can only exist one $Y$ value for an $X$ value by showing there exists at most one limit given $x_0$ is a cluster point.

Answer 1

Suppose $L_1$ and $L_2$ would be two different limits. Let $\varepsilon =\frac{|L_1 - L_2|}{2}$.

Apply the definitions of limit of $L_1$ and $L_2$ and get $\delta_1, \delta_2 >0$ from that. Then because $x$ is a limit point of $S$ we find some $x_1\in S$ with $0 < |x_1 - x_0| < \min(\delta_1,\delta_2)$.

Now derive the contradiction that

$$|L_1 - L_2| \le |L_1 - f(x_1)| + |L_2 - f(x_1)| < \varepsilon+ \varepsilon =|L_1-L_2|$$

So the limit is unique, if it exists.