LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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At a limit point of the domain a function has at most one limit

Statement

Let $A \subseteq R$ , let $f : A \to R$ , let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ) and let $L, L^{'} \in R$ . If

$\lim_{x \to c} f (x) = L and \lim_{x \to c} f (x) = L^{'}$

(The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ), then $L = L^{'}$ .

A function therefore has at most one limit at a limit point of its domain, which is what licenses the notation $\lim_{x \to c} f (x)$ for a single real number. This lemma is recorded in the justified_by field of The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ for exactly that reason.

The hypothesis that $c$ is a limit point is not removable. At an isolated point of the domain the same $ε$ - $δ$ formula is satisfied vacuously by every real at once, which is the content of FALSE: a function has at most one limit at every point of its domain, isolated points included.

Facts & Assumptions

Given: A set $A \subseteq R$ , a function $f : A \to R$ , a limit point $c$ of $A$ , and reals $L, L^{'}$ with $\lim_{x \to c} f (x) = L$ and $\lim_{x \to c} f (x) = L^{'}$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L1]

The limit condition: for every real $ε > 0$ there is a real $δ > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ$ satisfies $∣ f (x) - L ∣ < ε$ , and likewise with $L^{'}$ in place of $L$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

[L2]

Limit point: for every real $δ > 0$ there is $x \in A$ with $0 < ∣ x - c ∣ < δ$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

Triangle inequality: $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ in $R$ (The triangle inequality).

[L4]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ if and only if $u = 0$ ; and $∣ - u ∣ = ∣ u ∣$ (Basic properties of the absolute value).

[L5]

Order arithmetic in $R$ : trichotomy, so $u \neq 0$ together with $∣ u ∣ \geq 0$ and $∣ u ∣ \neq 0$ forces $∣ u ∣ > 0$ , and $t < t$ is impossible; adding two strict inequalities (Order is preserved by adding a constant and by adding inequalities); $0 < 1$ (The multiplicative identity is positive), hence $2 : = 1 + 1 > 0$ and $2^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order), so $η / 2 > 0$ and $(η / 2) \cdot 2 = η$ whenever $η > 0$ (Sign rules for products and monotonicity of multiplication, Ordered field); and of two positive reals the smaller is positive, the order being total.

Proof

technique · contradiction

1.1assume-contra

Suppose, for contradiction, that $L \neq L^{'}$ .

2.1step 1.1L4L5

Then $L - L^{'} \neq 0$ , so $∣ L - L^{'} ∣ \neq 0$ while $∣ L - L^{'} ∣ \geq 0$ , and trichotomy gives $∣ L - L^{'} ∣ > 0$ ; hence $ε : = ∣ L - L^{'} ∣ / 2 > 0$ and $2 ε = ∣ L - L^{'} ∣$ .

3.1step 2.1L1L5choose

Applying [L1] twice with this $ε$ , fix reals $δ_{1} > 0$ and $δ_{2} > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ_{1}$ has $∣ f (x) - L ∣ < ε$ and every $x \in A$ with $0 < ∣ x - c ∣ < δ_{2}$ has $∣ f (x) - L^{'} ∣ < ε$ ; put $δ$ to be the smaller of $δ_{1}$ and $δ_{2}$ , so $δ > 0$ .

4.1step 3.1L2choose

Since $c$ is a limit point of $A$ , fix $x \in A$ with $0 < ∣ x - c ∣ < δ$ .

5.1step 3.1step 4.1L1

That $x$ satisfies $0 < ∣ x - c ∣ < δ_{1}$ and $0 < ∣ x - c ∣ < δ_{2}$ , hence both $∣ f (x) - L ∣ < ε$ and $∣ f (x) - L^{'} ∣ < ε$ .

6.1step 5.1L3L4L5

Therefore $∣ L - L^{'} ∣ = ∣ (L - f (x)) + (f (x) - L^{'}) ∣ \leq ∣ L - f (x) ∣ + ∣ f (x) - L^{'} ∣ = ∣ f (x) - L ∣ + ∣ f (x) - L^{'} ∣ < ε + ε = 2 ε = ∣ L - L^{'} ∣$ .

7.1step 6.1L5discharge-contradiction∎

So $∣ L - L^{'} ∣ < ∣ L - L^{'} ∣$ , which trichotomy forbids; the assumption $L \neq L^{'}$ is untenable, and hence $L = L^{'}$ .

Remarks

Where each hypothesis is spent. The limit conditions are used only in step 5.1, and the limit-point hypothesis only in step 4.1, to produce a single point $x$ of the domain near $c$ at which both estimates can be read. Without such a point the two estimates never meet and nothing forces $L = L^{'}$ ; that is the whole mechanism, and it is the reason FALSE: a function has at most one limit at every point of its domain, isolated points included is false.
The sequential analogue is A sequence has at most one limit, proved by the same two-estimates-at-one-index argument. Neither statement uses any choice principle.
One-sided limits inherit this. By The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ a one-sided limit is the limit of a restriction of $f$ , so applying this lemma to that restriction gives uniqueness there too; nothing has to be reproved.

At a limit point of the domain a function has at most one limit

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources