LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The limit at $c$ depends only on the restriction of $f$ to a punctured neighbourhood of $c$ , and passes to any subset of the domain having $c$ as a limit point

Statement

Let $A \subseteq \mathbb{R}$ and let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

Locality. Let $f, g : A \to \mathbb{R}$ and $L \in \mathbb{R}$ , and suppose there is a real $\eta > 0$ with $f(x) = g(x)$ for every $x \in A$ satisfying $0 < |x - c| < \eta$ . Then $\lim_{x \to c} f(x) = L \iff \lim_{x \to c} g(x) = L$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ).
Restriction. Let $B \subseteq A$ with $c$ a limit point of $B$ , let $f : A \to \mathbb{R}$ and suppose $\lim_{x \to c} f(x) = L$ . Then $c$ is a limit point of $A$ as well, and $\lim_{x \to c} f|_B(x) = L$ , where $f|_B : B \to \mathbb{R}$ is the restriction of $f$ .

So the limit at $c$ sees only the values of $f$ on an arbitrarily small punctured neighbourhood of $c$ , and it survives shrinking the domain, provided the smaller domain still accumulates at $c$ . Together with At a limit point of the domain a function has at most one limit this is what makes the phrase the limit at $c$ a local notion.

The converse of claim 2 is false in general: a restriction may have a limit where the function has none, as the one-sided limits of the sign function on the companion page show.

Facts & Assumptions

Given: A set $A \subseteq \mathbb{R}$ and a limit point $c$ of $A$ ; for claim 1 functions $f, g : A \to \mathbb{R}$ , a real $L$ and a real $\eta > 0$ with $f(x) = g(x)$ for every $x \in A$ satisfying $0 < |x - c| < \eta$ ; for claim 2 a subset $B \subseteq A$ having $c$ as a limit point, a function $f : A \to \mathbb{R}$ and a real $L$ with $\lim_{x \to c} f(x) = L$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L1]

The limit condition: $\lim_{x \to c} h(x) = L$ means that for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that every $x$ in the domain of $h$ with $0 < |x - c| < \delta$ satisfies $|h(x) - L| < \varepsilon$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ).

[L2]

Limit point: $c$ is a limit point of a set $S$ when for every real $\delta > 0$ there is $x \in S$ with $0 < |x - c| < \delta$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L3]

Order arithmetic: of two positive reals the smaller is positive, the order being total; and $u < v \le w$ gives $u < w$ (Ordered field).

[L4]

Absolute value (Basic properties of the absolute value); and uniqueness of the limit at a limit point (At a limit point of the domain a function has at most one limit), which is what makes the phrase "the limit" in the statement denote.

Proof

technique · direct

1.1

For claim 1, assume $\lim_{x \to c} f(x) = L$ and let $\varepsilon > 0$ be an arbitrary real.

assume-hypL1

1.2

For claim 2, $B \subseteq A$ and $c$ is a limit point of $B$ ; hence $c$ is a limit point of $A$ , since for every real $\delta > 0$ a point $x \in B$ with $0 < |x - c| < \delta$ is also a point of $A$ with $0 < |x - c| < \delta$ .

1.3

For claim 2, assume $\lim_{x \to c} f(x) = L$ and let $\varepsilon > 0$ be an arbitrary real.

assume-hypL1

2.1

By [L1] fix a real $\delta_0 > 0$ such that every $x \in A$ with $0 < |x - c| < \delta_0$ satisfies $|f(x) - L| < \varepsilon$ , and put $\delta$ to be the smaller of $\delta_0$ and $\eta$ , so $\delta > 0$ .

step 1.1L1L3choose

2.2

By [L1] fix a real $\delta > 0$ such that every $x \in A$ with $0 < |x - c| < \delta$ satisfies $|f(x) - L| < \varepsilon$ .

step 1.3L1choose

3.1

Every $x \in A$ with $0 < |x - c| < \delta$ satisfies both $0 < |x - c| < \delta_0$ and $0 < |x - c| < \eta$ , so $g(x) = f(x)$ and $|g(x) - L| = |f(x) - L| < \varepsilon$ ; as $\varepsilon > 0$ was arbitrary, $\lim_{x \to c} g(x) = L$ .

step 2.1L1L3L4

3.2

Every $x \in B$ with $0 < |x - c| < \delta$ lies in $A$ and satisfies $0 < |x - c| < \delta$ , so $f|_B(x) = f(x)$ and therefore $|f(x) - L| < \varepsilon$ ; as $\varepsilon > 0$ was arbitrary, and $c$ is a limit point of $B$ , $\lim_{x \to c} f|_B(x) = L$ .

step 2.2L1L4

4.1

The hypothesis of claim 1 is symmetric in $f$ and $g$ , so interchanging their roles in steps 1.1, 2.1 and 3.1 gives the implication in the other direction, and claim 1 is proved; claim 2 is steps 1.2 and 3.2.

step 1.2step 3.1step 3.2∎

Remarks

What claim 1 is used for. It is the licence to modify a function outside a punctured neighbourhood of $c$ , or at $c$ itself, without changing the limit; the change at $c$ alone is already invisible to The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ , since the condition $0 < |x - c|$ excludes that point.
What claim 2 is used for. It is the step that lets a statement proved on $A$ be transported to a smaller domain: the one-sided limits of The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (-\infty, c)$ and $A \cap (c, \infty)$ are exactly limits of restrictions, and the quotient case of Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero is proved on the smaller domain where the denominator does not vanish.
Both claims are choice free. Only the $\varepsilon$ - $\delta$ definition is used; no sequence is constructed anywhere in the proof.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 33 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, §3.1: Limits of functions (standard reference, not scraped)
Limit of a function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.1 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §9.3 (standard reference, not scraped)

The limit at ccc depends only on the restriction of fff to a punctured neighbourhood of ccc, and passes to any subset of the domain having ccc as a limit point