The Uniqueness of Limits of a Function Theorem

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

Recall from The Limit of a Function page that for a function $f : A \to \mathbb{R}$ where $c$ is a cluster point of $A$, then $\lim_{x \to c} f(x) = L$ if $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $\mid f(x) - L \mid < \epsilon$. We have not yet established that the limit $L$ is unique, so is it possible that $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} f(x) = M$ where $L \neq M$? The following theorem will show that this cannot happen.
 Theorem (Uniqueness of Limits): Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then if $L, M \in \mathbb{R}$ are both limits of $f$ at $c$, that is $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} f(x) = M$, then $L = M$.
• Proof: Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Also let $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} f(x) = M$. Suppose that $L \neq M$. We will show that this leads to a contradiction. Let $\epsilon > 0$ be given.
• Since $\lim_{x \to c} f(x) = L$, then for $\epsilon_1 = \frac{\epsilon}{2}$ $\exists \delta_1 > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta_1$ then $\mid f(x) - L \mid < \epsilon_1 = \frac{\epsilon}{2}$.
• Similarly, since $\lim_{x \to c} f(x) = M$ then for $\epsilon_2 = \frac{\epsilon}{2}$ $\exists \delta_2 > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta_2$ then $\mid f(x) - M \mid < \epsilon_2 = \frac{\epsilon}{2}$. Now let $\delta = \mathrm{min} \{ \delta_1, \delta_2 \}$ and so we have that:
• But $\epsilon > 0$ is arbitrary, so this implies that $\mid L - M \mid = 0$, that is $L = M$, a contradiction. So our assumption that $L \neq M$ was false, and so if $\lim_{x \to c} f(x) = L$ then $L$ is unique. $\blacksquare$