The Limit Theorems for Functions

The Uniqueness of Limits of a Function Theorem

Recall from <a href="/the-limit-of-a-function">The Limit of a Function</a> 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:

${\begin{aligned}\quad \quad \mid L-M\mid =\mid L-f(x)+f(x)-M\mid \leq \mid L-f(x)\mid +\mid f(x)-M\mid <\epsilon _{1}+\epsilon _{2}={\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon \end{aligned}}$

• 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. $\blacksquare$