The Limit Theorems for Functions

The Uniqueness of Limits of a Function Theorem

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:

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

• But Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon > 0} is arbitrary, so this implies that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid L - M \mid = 0} , that is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L = M} , a contradiction. So our assumption that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L \neq M} was false, and so if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x \to c} f(x) = L} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} is unique. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}