# 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 $x \to x_0$

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.

• There's an error in your definition of limit. Let $L_1$ and $L_2$ be two limits. Then apply the definition of limit to each L. – GNUSupporter 8964民主女神 地下教會 Nov 24 '18 at 18:19

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 $$0 < |x_1 - x_0| < \min(\delta_1,\delta_2)$$.
$$|L_1 - L_2| \le |L_1 - f(x_1)| + |L_2 - f(x_1)| < \varepsilon+ \varepsilon =|L_1-L_2|$$