Proof of Uniqueness of the Limit of a Real Function at a Point of an Interval

1. (Punctured neighbourhoods are nonempty.) Let $\delta>0$. Since $I$ contains at least two points, it contains some $x_1$ with $x_1\ne c$, and then $|x_1-c|>0$. Put

so that $0<t$; moreover $t<\delta$ and $t<|x_1-c|$, since the halving of a positive element is positive and strictly smaller than it, by clause 8 of Elementary Order Arithmetic in an Ordered Field, and $\min\{\delta,|x_1-c|\}$ is at most each of $\delta$ and $|x_1-c|$.

Let $x=c+t$ if $c<x_1$, and $x=c-t$ if $x_1<c$; one of these cases holds because $x_1\ne c$ and the order of $\mathbb{R}$ is total. In either case $t<|x_1-c|$ places $x$ strictly between $c$ and $x_1$. Since $c\in I$ and $x_1\in I$, a point lying strictly between them belongs to $I$ by The Real Line: Standing Notation and Background for Calculus §intervals, so $x\in I$. Finally $|x-c|=t$, so $0<|x-c|<\delta$. Hence $x$ belongs to $\{x\in I:0<|x-c|<\delta\}$, which is therefore nonempty.

2. (Uniqueness.) Let $\varepsilon>0$. Choose $\delta>0$ for $L$ and $\delta'>0$ for $L'$ as in the hypothesis, and pick

which is possible by claim 1. Then $|f(x)-L|<\varepsilon$ and $|f(x)-L'|<\varepsilon$, so by the triangle inequality of Properties of the Absolute Value in an Ordered Field,

Suppose $|L-L'|>0$. Taking $\varepsilon=|L-L'|\cdot 2^{-1}$, which is positive by clause 8 of Elementary Order Arithmetic in an Ordered Field, the display gives $|L-L'|<|L-L'|$, which is impossible. Hence $|L-L'|=0$, and therefore $L=L'$ by Properties of the Absolute Value in an Ordered Field.

Consequently at most one real number satisfies the condition of Limit of a Real Function at a Point of an Interval §limit for $f$ at $c$, so the notation $\lim_{x\to c}f(x)$ introduced there is unambiguous.