Alphabeta Math
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.

✓ 17 results · all verified · 14 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Limits of Real Functions

1 · Prerequisites

2 · Summary

Objective. This page defines what it means for a function of a real variable to have a limit at a point, and proves the toolkit that makes the notion usable: uniqueness, locality, the algebra of limits, order preservation, the squeeze theorem, the relation between the two-sided limit and the two one-sided limits, and composition. It is built on the topology of R (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) and on the theory of sequences, and it is the last page before continuity.

The definition, and the three decisions inside it. The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A says that lim⁡x→cf(x)=L when for every real ε>0 there is a real δ>0 with ∣f(x)−L∣<ε for every x in the domain satisfying 0<∣x−c∣<δ. Three features are load bearing rather than decorative, and two of the three have a false statement on this page attached to them.

  • c must be a limit point of the domain. That is what keeps the quantified set nonempty for every δ, and hence what allows the condition to pin L down. At a limit point of the domain a function has at most one limit then proves that at most one L can occur, which is exactly what licenses the notation, and it is recorded in the definition's justified_by for that reason. Drop the hypothesis and uniqueness fails completely: at an isolated point every real satisfies the formula vacuously, which is FALSE: a function has at most one limit at every point of its domain, isolated points included. So at an isolated point of the domain the symbol lim⁡x→cf(x) is simply not defined here.
  • c need not lie in the domain, so a limit may be taken where the function is not defined.
  • The value f(c), when it exists, is invisible, because 0<∣x−c∣ removes x=c from the quantifier. Equality of limit and value is therefore a hypothesis and not a consequence, which is FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist; that equality is what continuity at c will mean, on the next page of this track.

Locality. 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 proves the two statements that make the limit a local object: changing f outside a punctured neighbourhood of c changes nothing, and a limit survives restricting the domain to any subset that still has c as a limit point. Everything later on this page that shrinks a domain — the one-sided limits, the quotient rule — goes through it.

The two variants. The left and right limits of f at c, as limits of the restrictions of f to A∩(−∞,c) and A∩(c,∞) defines lim⁡x→c±f(x) as the limit of the restriction of f to the points of the domain on one side of c, so uniqueness and locality are inherited rather than reproved. Limits at +∞ and −∞, and infinite limits at a point defines limits at ±∞, where the role of the limit-point hypothesis is played by unboundedness of the domain, and infinite limits at a point. The symbols ±∞ remain abbreviations and never real numbers: the library does not write lim⁡x→cf(x)=+∞, for the reason Divergence to +∞ and to −∞ already gave for sequences. Uniqueness of the limit at ±∞ is proved inside that definition.

Choice hygiene, and why it shapes the page. Heine criterion: lim⁡x→cf(x)=L iff f(xk)→L for every sequence in A∖{c} converging to c — the Heine criterion — says that lim⁡x→cf(x)=L if and only if f(xk)→L for every sequence in the domain avoiding c and converging to c. Its two directions do not cost the same. The direction from ε-δ to sequences is a theorem of ZF. The converse, as proved here, invokes the axiom of countable choice (The Axiom of Countable Choice (ACω)) exactly once, to pick one bad point from each of countably many nonempty sets — the same use, for the same reason, as in A point lies in the closure of A⊆R iff some sequence in A converges to it, so a subset of R is closed iff it is sequentially closed on the prerequisite page. The sequence-to-ε direction of the Heine criterion uses countable choice for R, and where this library records that cost records precisely what is and is not claimed about that: in particular this library does not claim the axiom is necessary, and it warns against the slogan that sequential criteria always need choice, since Sierpiński's theorem on everywhere-sequentially-continuous functions is a theorem of ZF.

The consequence for this page is a deliberate organisation. Everything that can be proved directly from ε and δ is proved that way, so the algebra of limits, order preservation, the squeeze theorem and composition are all theorems of ZF. The sequential side is used only where it earns its place: in the criterion itself, and in A function has no limit at c as soon as two sequences in A∖{c} tending to c give different limits of the values, which needs only the choice-free direction and is the standard way to show that a limit does not exist — two sequences tending to c whose image sequences tend to different values, or one whose image sequence does not converge at all.

The toolkit, in the order it is needed. If f has a finite limit at c then f is bounded on some punctured neighbourhood of c shows that a function with a limit at c is bounded on some punctured neighbourhood of c — and only there, which is FALSE: a function with a limit at c is bounded on its whole domain. If lim⁡x→cf(x)=L≠0 then ∣f∣>∣L∣/2 on a punctured neighbourhood of c; in particular if L>0 then f>L/2>0 there shows that a nonzero limit L forces ∣f∣>∣L∣/2 on a punctured neighbourhood, with the sign of L, and that c remains a limit point of the set where f does not vanish. Those two lemmas are exactly what the product and quotient cases of the next theorem need, which is why they precede it.

Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero then proves that sums, scalar multiples, products and quotients of limits behave as expected, the quotient on the domain where the denominator does not vanish and under the hypothesis that its limit is nonzero. Each claim asserts both that the compound limit exists and what it equals. If f≤g on a punctured neighbourhood of c then lim⁡f≤lim⁡g, non-strictly proves that f≤g near c gives lim⁡f≤lim⁡g; the conclusion cannot be sharpened to a strict inequality even from a strict hypothesis, which is FALSE: f<g near c implies lim⁡f<lim⁡g. If f≤g≤h near c and f and h have the same limit at c, then so does g is the one result here that produces a limit rather than computing one: no hypothesis is placed on the squeezed function at all. If c is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree closes the loop with the one-sided limits.

Composition, with the hypothesis that is usually left implicit. Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c is false as usually first stated. The inner limit controls ∣g(x)−L∣ but does not prevent g(x) from equalling L, and at those arguments the outer limit says nothing, since it never sees f(L). Two hypotheses each close the gap, and either suffices: (i) L lies in the outer domain with f(L)=M, which is continuity of f at L written out; or (ii) g avoids the value L on a punctured neighbourhood of c, which is what makes substitutions such as y=1/x legitimate. With both dropped the statement is refuted by FALSE: lim⁡x→cf(g(x))=M whenever lim⁡x→cg=L and lim⁡y→Lf=M, whose witness fails (i) and (ii) at once.

One reusable lemma, deliberately placed here. Integer part: for every real x there is exactly one integer m with m≤x<m+1 proves that every real x has exactly one integer m with m≤x<m+1. Existence is the Archimedean property together with the well-ordering of N; uniqueness is the discreteness of Z. It is the library's first floor item, and it is stated on this page rather than inside an example so that later pages — monotone functions, powers, content — can cite it instead of rebuilding the argument. Its immediate use is on the companion page, where it computes the trigonometry-free oscillator ψ(x)=inf⁡n∈Z∣x−n∣ in one line.

The companion page carries the witnesses: polynomials and rational functions, the oscillator ψ and the two examples built on it, the sign function, a limit at +∞ computed by a direct estimate, the indicator of Q, and the counterexample items that work three of the five false statements listed here out in full. Each of the five already carries its own witness, verified in the false statement itself; what the companion page adds, for those three, is the further computation each witness supports.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A

Definition

Throughout, R is the complete ordered field (Complete ordered field (least-upper-bound property)) with its order and absolute value (Order on the reals).

Let A⊆R, let f:A→R, let c∈R be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), and let L∈R. We say that f(x) tends to L as x tends to c, and write

lim⁡x→cf(x)=L,

when

(∀ε>0) (∃δ>0) (∀x∈A) [ 0<∣x−c∣<δ ⟹ ∣f(x)−L∣<ε ],

where ε and δ range over the positive reals.

In the language of neighbourhoods (The ε-neighbourhood and the punctured ε-neighbourhood of a point of R) the condition reads: for every real ε>0 there is a real δ>0 with

f(A∩Nδ∗(c))  ⊆  Nε(L),

Nδ∗(c)={ y:0<∣y−c∣<δ } being the punctured δ-neighbourhood of c and Nε(L)=(L−ε, L+ε) the open interval of Intervals of R: the nine order-convex forms, nondegeneracy, and length. The two forms agree because ∣f(x)−L∣<ε says exactly f(x)∈Nε(L), and 0<∣x−c∣<δ says exactly x∈Nδ∗(c).

Three features of this definition are load bearing, not decoration.

  1. c is required to be a limit point of A. By Limit point, isolated point, adherent point, derived set, and dense subset of R that says every punctured neighbourhood of c meets A, so for every δ>0 the set A∩Nδ∗(c) over which the implication quantifies is nonempty. Drop the requirement and the implication can be satisfied vacuously by every real L at once, which is exactly what FALSE: a function has at most one limit at every point of its domain, isolated points included records. At a point of A that is not a limit point of A — an isolated point — the symbol lim⁡x→cf(x) is therefore not defined in this library.

  2. c∈A is not required. A limit point of A need not belong to A (Limit point, isolated point, adherent point, derived set, and dense subset of R), and the definition never evaluates f at c. This is what allows a limit to be taken at a point where the function is not defined at all, as at 0 for x↦x ψ(1/x).

  3. The value f(c), when it exists, is irrelevant. The hypothesis 0<∣x−c∣ excludes x=c from the quantifier, so changing f at the single point c changes nothing. Equality of the limit with the value is an extra condition, not a consequence: FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist.

The notation presumes uniqueness. Writing lim⁡x→cf(x)=L treats the left-hand side as a name for a single real number, which is legitimate only because at a limit point at most one L can satisfy the displayed condition. That obligation is discharged by At a limit point of the domain a function has at most one limit ↗, recorded in this item's justified_by. As with sup⁡S (Conventions: sup⁡∅, unbounded sets, and the extended reals) and lim⁡kxk (A sequence has at most one limit), the symbol is written only for a function already known to have a limit at c.

Real and rational ε define the same relation. Above, ε and δ range over the positive reals. Restricting either quantifier to the positive rationals gives the same relation: every positive rational is a positive real, and below every positive real lies a positive rational (The rationals embed densely in the reals), so an ε-condition verified for all positive rationals is verified for an arbitrary positive real η by running it at a rational ε with 0<ε<η, and a δ produced as a real may be shrunk to a rational one below it. This is the passage sanctioned in the remarks of Sequences of reals: bounded, eventually, frequently, tails, subsequences, and it is what lets this definition be compared with Limits and Cauchy sequences of reals, whose ε is rational, in Heine criterion: lim⁡x→cf(x)=L iff f(xk)→L for every sequence in A∖{c} converging to c.

Remarks

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

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

Statement

Let A⊆R, let f:A→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′∈R. If

lim⁡x→cf(x)=Landlim⁡x→cf(x)=L′

(The ε-δ limit lim⁡x→cf(x)=L of f:A→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→cf(x) for a single real number. This lemma is recorded in the justified_by field of The ε-δ limit lim⁡x→cf(x)=L of f:A→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⊆R, a function f:A→R, a limit point c of A, and reals L,L′ with lim⁡x→cf(x)=L and lim⁡x→cf(x)=L′ (The ε-δ limit lim⁡x→cf(x)=L of f:A→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∈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→cf(x)=L of f:A→R at a limit point c of A).

[L2]
[L3]

Triangle inequality: ∣u+v∣≤∣u∣+∣v∣ in R (The triangle inequality).

[L4]

Absolute value: ∣u∣≥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≠0 together with ∣u∣≥0 and ∣u∣≠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)⋅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
1.1

Suppose, for contradiction, that L≠L′.

assume-contra
2.1

Then L−L′≠0, so ∣L−L′∣≠0 while ∣L−L′∣≥0, and trichotomy gives ∣L−L′∣>0; hence ε:=∣L−L′∣/2>0 and 2ε=∣L−L′∣.

step 1.1L4L5
3.1

Applying [L1] twice with this ε, fix reals δ1>0 and δ2>0 such that every x∈A with 0<∣x−c∣<δ1 has ∣f(x)−L∣<ε and every x∈A with 0<∣x−c∣<δ2 has ∣f(x)−L′∣<ε; put δ to be the smaller of δ1 and δ2, so δ>0.

step 2.1L1L5choose
4.1

Since c is a limit point of A, fix x∈A with 0<∣x−c∣<δ.

step 3.1L2choose
5.1

That x satisfies 0<∣x−c∣<δ1 and 0<∣x−c∣<δ2, hence both ∣f(x)−L∣<ε and ∣f(x)−L′∣<ε.

step 3.1step 4.1L1
6.1

Therefore ∣L−L′∣=∣(L−f(x))+(f(x)−L′)∣≤∣L−f(x)∣+∣f(x)−L′∣=∣f(x)−L∣+∣f(x)−L′∣<ε+ε=2ε=∣L−L′∣.

step 5.1L3L4L5
7.1

So ∣L−L′∣<∣L−L′∣, which trichotomy forbids; the assumption L≠L′ is untenable, and hence L=L′.

step 6.1L5discharge-contradiction∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

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⊆R and let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R).

  1. Locality. Let f,g:A→R and L∈R, and suppose there is a real η>0 with f(x)=g(x) for every x∈A satisfying 0<∣x−c∣<η. Then lim⁡x→cf(x)=L  ⟺  lim⁡x→cg(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

  2. Restriction. Let B⊆A with c a limit point of B, let f:A→R and suppose lim⁡x→cf(x)=L. Then c is a limit point of A as well, and lim⁡x→cf∣B(x)=L, where f∣B:B→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⊆R and a limit point c of A; for claim 1 functions f,g:A→R, a real L and a real η>0 with f(x)=g(x) for every x∈A satisfying 0<∣x−c∣<η; for claim 2 a subset B⊆A having c as a limit point, a function f:A→R and a real L with lim⁡x→cf(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→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: lim⁡x→ch(x)=L means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−L∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Limit point: c is a limit point of a set S when for every real δ>0 there is x∈S 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]

Order arithmetic: of two positive reals the smaller is positive, the order being total; and u<v≤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→cf(x)=L and let ε>0 be an arbitrary real.

assume-hypL1
1.2

For claim 2, B⊆A and c is a limit point of B; hence c is a limit point of A, since for every real δ>0 a point x∈B with 0<∣x−c∣<δ is also a point of A with 0<∣x−c∣<δ.

L2
1.3

For claim 2, assume lim⁡x→cf(x)=L and let ε>0 be an arbitrary real.

assume-hypL1
2.1

By [L1] fix a real δ0>0 such that every x∈A with 0<∣x−c∣<δ0 satisfies ∣f(x)−L∣<ε, and put δ to be the smaller of δ0 and η, so δ>0.

step 1.1L1L3choose
2.2

By [L1] fix a real δ>0 such that every x∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε.

step 1.3L1choose
3.1

Every x∈A with 0<∣x−c∣<δ satisfies both 0<∣x−c∣<δ0 and 0<∣x−c∣<η, so g(x)=f(x) and ∣g(x)−L∣=∣f(x)−L∣<ε; as ε>0 was arbitrary, lim⁡x→cg(x)=L.

step 2.1L1L3L4
3.2

Every x∈B with 0<∣x−c∣<δ lies in A and satisfies 0<∣x−c∣<δ, so f∣B(x)=f(x) and therefore ∣f(x)−L∣<ε; as ε>0 was arbitrary, and c is a limit point of B, lim⁡x→cf∣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

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The left and right limits of f at c, as limits of the restrictions of f to A∩(−∞,c) and A∩(c,∞)

Definition

Let A⊆R, let f:A→R and let c∈R. Put

A−:=A∩(−∞,c),A+:=A∩(c,∞)

(Intervals of R: the nine order-convex forms, nondegeneracy, and length), and write f−:=f∣A− and f+:=f∣A+ for the restrictions of f to those sets.

Right limit. Suppose c is a limit point of A+ (Limit point, isolated point, adherent point, derived set, and dense subset of R). For L∈R we write

lim⁡x→c+f(x)=L:⟺lim⁡x→cf+(x)=L

in the sense of The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A. Written out: for every real ε>0 there is a real δ>0 such that

∣f(x)−L∣<εfor every x∈A with c<x<c+δ.

Left limit. Suppose c is a limit point of A−. For L∈R we write lim⁡x→c−f(x)=L when lim⁡x→cf−(x)=L; written out, for every real ε>0 there is a real δ>0 with ∣f(x)−L∣<ε for every x∈A with c−δ<x<c.

The written-out forms agree with the definitions. For x∈A+ the two conditions 0<∣x−c∣<δ and c<x<c+δ are the same: x>c gives x−c>0, so ∣x−c∣=x−c and 0<∣x−c∣<δ reads 0<x−c<δ (Basic properties of the absolute value). Symmetrically on the left, where x<c gives ∣x−c∣=c−x.

Well-posedness is inherited, not reproved. A one-sided limit is a limit, namely the limit of a restriction, so:

When the symbols are defined. If c is not a limit point of A+ — for instance if A contains no point to the right of c, or only points bounded away from c on that side — then lim⁡x→c+f(x) is not defined here, for the reason given in The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A: the ε-δ condition would be satisfied vacuously by every real at once. The same applies on the left.

Remarks

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Limits at +∞ and −∞, and infinite limits at a point

Definition

Throughout, +∞ and −∞ are abbreviations and not real numbers, exactly as in Intervals of R: the nine order-convex forms, nondegeneracy, and length and Divergence to +∞ and to −∞. Every phrase below is a single abbreviation for a displayed condition on reals, and no arithmetic is ever performed with the symbols.

Limits at +∞. Let A⊆R be not bounded above (Lower bound, bounded below, bounded set), let f:A→R and let L∈R. We write

lim⁡x→+∞f(x)=L

when for every real ε>0 there is a real M such that

∣f(x)−L∣<εfor every x∈A with x>M.

Limits at −∞. Let A be not bounded below. We write lim⁡x→−∞f(x)=L when for every real ε>0 there is a real M with ∣f(x)−L∣<ε for every x∈A with x<M.

Why unboundedness is required. It plays exactly the role the limit-point condition plays in The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A. Saying that A is not bounded above says that no real is an upper bound of A, that is, that for every real M there is x∈A with x>M (Lower bound, bounded below, bounded set, Complete ordered field (least-upper-bound property)); so the set over which the condition quantifies is never empty and the condition is never vacuous. Without the hypothesis every real L would satisfy it and the notation would not denote.

Uniqueness, proved here. Suppose A is not bounded above and lim⁡x→+∞f(x)=L and lim⁡x→+∞f(x)=L′ with L≠L′. Then ∣L−L′∣>0 (Basic properties of the absolute value), so ε:=∣L−L′∣/2>0 (The multiplicative identity is positive, Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication). Choose reals M1,M2 witnessing the two conditions at this ε and let M be the larger of them, the order being total. Since A is not bounded above there is x∈A with x>M, hence with x>M1 and x>M2, and then

∣L−L′∣=∣(L−f(x))+(f(x)−L′)∣≤∣f(x)−L∣+∣f(x)−L′∣<2ε=∣L−L′∣

(The triangle inequality, Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities), which trichotomy forbids. So L=L′, and the notation lim⁡x→+∞f(x) denotes a single real. The same four lines, with the inequalities on x reversed, give uniqueness at −∞.

Infinite limits at a point. Let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R) and let f:A→R. We write

f(x)→+∞  as  x→c

when for every real M there is a real δ>0 such that f(x)>M for every x∈A with 0<∣x−c∣<δ; and f(x)→−∞ as x→c when for every real M there is a real δ>0 with f(x)<M for every such x.

This library does not write lim⁡x→cf(x)=+∞. The right-hand side would not be an element of R, and writing the equation would silently move the discussion into the extended real line, a structure that is not a field. That is the convention already fixed by Divergence to +∞ and to −∞ for sequences and by Conventions: sup⁡∅, unbounded sets, and the extended reals for suprema, and it is kept here. In particular none of the rules of Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero may be applied to a function tending to ±∞.

Combined forms. Let A be not bounded above and f:A→R. We write f(x)→+∞ as x→+∞ when for every real N there is a real M with f(x)>N for every x∈A with x>M. The other forms are obtained the same way, by pairing one of the two conditions on x (unbounded above, unbounded below) with one of the two conditions on f(x) (above every real, below every real); each is again a single abbreviation for the displayed condition, and none of them is an equation.

Remarks

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

Heine criterion: lim⁡x→cf(x)=L iff f(xk)→L for every sequence in A∖{c} converging to c

Statement

Let A⊆R, let f:A→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∈R. The following are equivalent.

  1. lim⁡x→cf(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).
  2. For every sequence (xk)k∈N with xk∈A and xk≠c for every k, and xk→c (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), the sequence (f(xk))k∈N converges to L.

The two directions do not cost the same. The implication from 1 to 2 is proved in ZF: the sequence is handed to the proof, and nothing is selected. The implication from 2 to 1, as proved below, invokes the axiom of countable choice (The Axiom of Countable Choice (ACω)) exactly once, at step 3.2, to select one bad point from each of countably many nonempty sets. What this library does and does not claim about that cost is recorded in The sequence-to-ε direction of the Heine criterion uses countable choice for R, and where this library records that cost; the same asymmetry appears, for the same reason, in A point lies in the closure of A⊆R iff some sequence in A converges to it, so a subset of R is closed iff it is sequentially closed.

Because of this, the results on this page that can be proved directly from ε and δ — the algebra of limits, order preservation, the squeeze theorem, composition — are proved that way, and not through this criterion. What the criterion is for is the transfer of sequential results to functions, and above all the negative use recorded in A function has no limit at c as soon as two sequences in A∖{c} tending to c give different limits of the values, which needs only the choice-free direction.

Facts & Assumptions

Given: A set A⊆R, a function f:A→R, a limit point c of A and a real L. Sequences are functions on N, and N contains 0 (Sequences of reals: bounded, eventually, frequently, tails, subsequences, The natural numbers N (von Neumann)), so the shrinking radii used below are 1/(k+1) and never 1/k.

[L1]

The function limit: lim⁡x→cf(x)=L means that for every real ε>0 there is a real δ>0 such that every x∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Sequential convergence: (yk)→y means that for every rational ε>0 there is K∈N with ∣yk−y∣<ε for all k≥K (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences). Testing instead against every positive REAL ε defines the same relation: every positive rational is a positive real, and below every positive real lies a positive rational (The rationals embed densely in the reals), which is the passage sanctioned in the remarks of Sequences of reals: bounded, eventually, frequently, tails, subsequences.

[L3]
[L4]

Reciprocal Archimedean property: for every real ε>0 there is a natural n≥1 with 1/n<ε (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε, Every complete ordered field is Archimedean); the canonical naturals satisfy n⋅1R>0 and are strictly increasing in n (Canonical naturals are positive and strictly increasing); and 0<a<b gives 0<1/b<1/a (Inverses of positives are positive, and reciprocation reverses order).

[L5]

Countable choice: for every family (Xk)k∈N of nonempty sets there is a function k↦xk with xk∈Xk for every k (The Axiom of Countable Choice (ACω)).

[L6]

Absolute value (Basic properties of the absolute value); and trichotomy, so the negation of ∣u∣<ε is ∣u∣≥ε, and the negation of "for every ε there is δ such that P" is "there is ε0 such that for every δ, not P" (Ordered field).

Proof

technique · direct
1.1

Assume condition 1, let (xk) be a sequence with xk∈A and xk≠c for every k and xk→c, and let ε>0 be an arbitrary real.

assume-hypL1L2L3
1.2

Assume condition 1 FAILS. Negating the quantifiers of [L1], there is a real ε0>0 such that for every real δ>0 some x∈A has 0<∣x−c∣<δ and ∣f(x)−L∣≥ε0.

assume-hypL1L6
2.1

By [L1] fix a real δ>0 such that every x∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε; and by [L2], δ being a positive real, fix K∈N with ∣xk−c∣<δ for every k≥K.

step 1.1L1L2choose
2.2

For k∈N put Xk:={ x∈A : 0<∣x−c∣<1/(k+1)  and  ∣f(x)−L∣≥ε0 }. Each Xk is nonempty, since k+1≥1 makes 1/(k+1) a positive real and step 1.2 applies to that radius.

step 1.2L4L6
3.1

For every k≥K we have xk∈A and xk≠c, so 0<∣xk−c∣<δ and hence ∣f(xk)−L∣<ε. Since ε>0 was an arbitrary real, f(xk)→L; condition 1 therefore implies condition 2.

step 2.1L1L2L6
3.2

By countable choice applied to the family (Xk)k∈N, fix a function k↦xk with xk∈Xk for every k∈N.

step 2.2L5choose
4.1

That sequence has xk∈A and xk≠c for every k, and it converges to c: given a real ε>0, [L4] supplies a natural n≥1 with 1/n<ε, and every k≥n has k+1>n≥1, hence ∣xk−c∣<1/(k+1)<1/n<ε.

step 3.2L2L4L6
4.2

Yet (f(xk)) does not converge to L: every k has ∣f(xk)−L∣≥ε0, while a rational ε with 0<ε<ε0 ([L2]) would require some K with ∣f(xk)−L∣<ε<ε0 for all k≥K.

step 3.2L2L6
5.1

So the failure of condition 1 produces a sequence witnessing the failure of condition 2; contrapositively, condition 2 implies condition 1, and with step 3.1 the two conditions are equivalent.

step 3.1step 4.1step 4.2∎

Remarks

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

A function has no limit at c as soon as two sequences in A∖{c} tending to c give different limits of the values

Statement

Let A⊆R, let f:A→R and let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R). Then f has no limit at c — that is, no L∈R satisfies lim⁡x→cf(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A) — as soon as either of the following occurs.

  1. There are sequences (xk) and (yk) with all terms in A∖{c}, both converging to c, and reals P≠Q with f(xk)→P and f(yk)→Q (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).
  2. There is a sequence (xk) with all terms in A∖{c}, converging to c, for which (f(xk)) does not converge.

Only the choice-free half of the Heine criterion is used. The proof runs the implication from condition 1 to condition 2 of Heine criterion: lim⁡x→cf(x)=L iff f(xk)→L for every sequence in A∖{c} converging to c, which is a theorem of ZF; no sequence is constructed here, both being supplied by the hypothesis. So this corollary, the workhorse for showing that a limit fails to exist, costs no choice principle at all.

Facts & Assumptions

[L1]

Heine criterion, the direction from the ε-δ limit to sequences: if lim⁡x→cf(x)=L then f(zk)→L for every sequence (zk) with all terms in A∖{c} converging to c (Heine criterion: lim⁡x→cf(x)=L iff f(xk)→L for every sequence in A∖{c} converging to c). That direction is proved without any choice principle.

[L2]

A sequence of reals has at most one limit, so two limits of the same sequence are equal, and a sequence with a limit converges (A sequence has at most one limit, Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Proof

technique · contrapositive
1.1

Each of the two claims has the form "hypothesis ⇒ f has no limit at c"; we prove the contrapositive of each, namely that if some L∈R satisfies lim⁡x→cf(x)=L then neither hypothesis can hold.

contrapositive-reduce
1.2

Assume there is L∈R with lim⁡x→cf(x)=L.

assume-hyp
2.1

Let (zk) be an arbitrary sequence with all terms in A∖{c} converging to c. By [L1], (f(zk)) converges, with limit L.

step 1.2L1
3.1

Under hypothesis 1 this applies to (xk) and to (yk): f(xk)→P and f(xk)→L give P=L by [L2], and likewise Q=L, so P=Q; hypothesis 1, which asserts P≠Q, therefore fails.

step 2.1L2
3.2

Under hypothesis 2 it applies to (xk) and gives that (f(xk)) converges; hypothesis 2, which asserts that it does not, therefore fails.

step 2.1L2
4.1

So the existence of a limit of f at c excludes both hypotheses; contrapositively, either hypothesis excludes the existence of a limit of f at c.

step 3.1step 3.2discharge-contrapositive∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

If f has a finite limit at c then f is bounded on some punctured neighbourhood of c

Statement

Let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), let f:A→R and suppose the limit of f at c exists, say lim⁡x→cf(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Then there are a real δ>0 and a real M≥0 with

∣f(x)∣≤Mfor every x∈A with 0<∣x−c∣<δ;

equivalently, the image f(A∩Nδ∗(c)) is a bounded subset of R (Lower bound, bounded below, bounded set, The ε-neighbourhood and the punctured ε-neighbourhood of a point of R). One may take M=∣L∣+1.

Only local boundedness follows, never boundedness on A. A function with a limit at c may be unbounded on its domain, as FALSE: a function with a limit at c is bounded on its whole domain records.

Facts & Assumptions

Given: A set A⊆R, a limit point c of A, a function f:A→R and a real L with lim⁡x→cf(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→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∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Absolute value: ∣u∣≥0; −∣u∣≤u≤∣u∣; and for t>0, ∣u∣≤t is equivalent to −t≤u≤t (Basic properties of the absolute value).

[L3]

Triangle inequality: ∣u+v∣≤∣u∣+∣v∣ (The triangle inequality).

[L4]

Order arithmetic: 0<1 (The multiplicative identity is positive); adding a constant preserves the order and adding inequalities is legitimate (Order is preserved by adding a constant and by adding inequalities); and u<v implies u≤v. Order is preserved by adding a constant and by adding inequalities states these moves in their STRICT forms only; the non-strict forms used below follow by adjoining the equality case, in which the two sides coincide, the order being total (Ordered field).

[L5]

Bounded set: S⊆R is bounded when it has both an upper and a lower bound (Lower bound, bounded below, bounded set); and Nδ∗(c)={ y:0<∣y−c∣<δ } (The ε-neighbourhood and the punctured ε-neighbourhood of a point of R).

Proof

technique · direct
1.1

Apply [L1] with the particular value ε=1, legitimate since 1>0: fix a real δ>0 such that every x∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<1.

L1L4choose
1.2

Put M:=∣L∣+1. Then M≥0, since ∣L∣≥0 and 1>0.

L2L4
2.1

For every x∈A with 0<∣x−c∣<δ we have ∣f(x)∣=∣(f(x)−L)+L∣≤∣f(x)−L∣+∣L∣<1+∣L∣=M, hence ∣f(x)∣≤M.

step 1.1step 1.2L2L3L4
3.1

Therefore −M≤f(x)≤M for every such x, so M is an upper bound and −M a lower bound of the image f(A∩Nδ∗(c)): that image is a bounded subset of R.

step 2.1L2L5∎

Remarks

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

If lim⁡x→cf(x)=L≠0 then ∣f∣>∣L∣/2 on a punctured neighbourhood of c; in particular if L>0 then f>L/2>0 there

Statement

Let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), let f:A→R and suppose the limit of f at c exists with lim⁡x→cf(x)=L and L≠0 (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Then there is a real δ>0 such that every x∈A with 0<∣x−c∣<δ satisfies

∣f(x)∣  >  ∣L∣2  >  0;

in particular f(x)≠0 for every such x. Moreover:

  • if L>0 then f(x)>L/2>0 for every such x;
  • if L<0 then f(x)<L/2<0 for every such x.

Consequently, writing

A0:={ x∈A : f(x)≠0 },

the point c is a limit point of A0.

The bound ∣L∣/2, and not merely "f≠0", is what later proofs need. The quotient case of Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero estimates 1/∣f∣ near c and therefore needs a positive lower bound on ∣f∣ there, and the last claim is what lets a limit be taken on the smaller domain A0 at all.

Facts & Assumptions

Given: A set A⊆R, a limit point c of A, a function f:A→R and a real L≠0 with lim⁡x→cf(x)=L; and A0:={ x∈A:f(x)≠0 } (The ε-δ limit lim⁡x→cf(x)=L of f:A→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∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Absolute value: ∣u∣≥0; ∣u∣=0 if and only if u=0; ∣u∣=u for u≥0 and ∣u∣=−u for u≤0; and for t>0, ∣u∣<t is equivalent to −t<u<t (Basic properties of the absolute value).

[L3]

Reverse triangle inequality: ∣∣u∣−∣v∣∣≤∣u−v∣ (The reverse triangle inequality).

[L4]
[L5]

Order arithmetic in R: trichotomy, so u≠0 with ∣u∣≥0 and ∣u∣≠0 forces ∣u∣>0; 0<1 (The multiplicative identity is positive), hence 2>0 and 2−1>0 (Inverses of positives are positive, and reciprocation reverses order), so t/2>0 and t−t/2=t/2 for t>0 (Sign rules for products and monotonicity of multiplication); adding a constant to an inequality (Order is preserved by adding a constant and by adding inequalities); and of two positive reals the smaller is positive, the order being total (Ordered field).

Proof

technique · direct
1.1

Since L≠0 we have ∣L∣≠0 while ∣L∣≥0, so trichotomy gives ∣L∣>0, and ε:=∣L∣/2>0 with ∣L∣−∣L∣/2=∣L∣/2.

givenL2L5
2.1

Apply [L1] with this ε: fix a real δ>0 such that every x∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<∣L∣/2.

step 1.1L1choose
3.1

For every such x the reverse triangle inequality gives ∣∣f(x)∣−∣L∣∣≤∣f(x)−L∣<∣L∣/2, hence ∣f(x)∣−∣L∣>−∣L∣/2 and so ∣f(x)∣>∣L∣−∣L∣/2=∣L∣/2>0; in particular ∣f(x)∣≠0 and therefore f(x)≠0.

step 2.1L2L3L5
3.2

If L>0 then ∣L∣=L, and for every such x the estimate ∣f(x)−L∣<L/2 gives −L/2<f(x)−L, that is f(x)>L−L/2=L/2>0.

step 2.1L2L5
3.3

If L<0 then ∣L∣=−L, and for every such x the estimate ∣f(x)−L∣<−L/2 gives f(x)−L<−L/2, that is f(x)<L−L/2=L/2<0.

step 2.1L2L5
4.1

Let η>0 be an arbitrary real and let ρ be the smaller of δ and η, so ρ>0. Since c is a limit point of A there is x∈A with 0<∣x−c∣<ρ; that x satisfies 0<∣x−c∣<δ, hence f(x)≠0 by step 3.1, so x∈A0 and 0<∣x−c∣<η. As η was arbitrary, c is a limit point of A0.

step 3.1L4L5
5.1

So on A∩Nδ∗(c) the function is bounded away from 0 by ∣L∣/2 and carries the sign of L, and c remains a limit point of the set A0 where f does not vanish.

step 3.1step 3.2step 3.3step 4.1∎

Remarks

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero

Statement

Let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), let f,g:A→R and let α∈R. Suppose the limits of f and of g at c exist, and write L:=lim⁡x→cf(x) and M:=lim⁡x→cg(x) (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Then:

  1. the limit of f+g at c exists, and lim⁡x→c(f+g)(x)  =  lim⁡x→cf(x)+lim⁡x→cg(x)  =  L+M;
  2. the limit of αf at c exists, and lim⁡x→c(αf)(x)  =  αlim⁡x→cf(x)  =  αL;
  3. the limit of fg at c exists, and lim⁡x→c(fg)(x)  =  (lim⁡x→cf(x))(lim⁡x→cg(x))  =  LM;
  4. if M≠0, then, writing A0:={ x∈A:g(x)≠0 }, the point c is a limit point of A0, the quotient f/g is defined on A0 by (f/g)(x)=f(x)/g(x), the limit of (f/g)∣A0 at c exists, and lim⁡x→c(f/g)∣A0(x)  =  lim⁡x→cf(x)lim⁡x→cg(x)  =  LM.

Each equation asserts two things at once: that the limit on the left exists, and that it has the stated value. Both are proved. The symbols denote by At a limit point of the domain a function has at most one limit.

Everything below is proved directly from ε and δ. No sequence is constructed and no choice principle is used, so all four claims are theorems of ZF. Passing through Heine criterion: lim⁡x→cf(x)=L iff f(xk)→L for every sequence in A∖{c} converging to c instead would import the countable choice spent in that theorem's converse direction, for no gain; see The sequence-to-ε direction of the Heine criterion uses countable choice for R, and where this library records that cost.

Why the quotient is stated on A0. The function f/g is simply not defined where g vanishes, and g may well vanish at points of A arbitrarily far from c; restricting to A0 is therefore forced. That this restriction still has c as a limit point, so that the limit there means anything at all, is the last claim of If lim⁡x→cf(x)=L≠0 then ∣f∣>∣L∣/2 on a punctured neighbourhood of c; in particular if L>0 then f>L/2>0 there. The sequential analogue Algebra of limits: sums, scalar multiples, products and quotients needs the corresponding hypothesis in the form "the denominator sequence is nonzero at every index".

Facts & Assumptions

Given: A set A⊆R, a limit point c of A, functions f,g:A→R, a real α, and reals L,M with lim⁡x→cf(x)=L and lim⁡x→cg(x)=M; for claim 4 also M≠0 and A0:={ x∈A:g(x)≠0 } (The ε-δ limit lim⁡x→cf(x)=L of f:A→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: lim⁡x→ch(x)=P means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−P∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Absolute value: ∣u∣≥0; ∣u∣=0 if and only if u=0; ∣uv∣=∣u∣ ∣v∣; and ∣−u∣=∣u∣ (Basic properties of the absolute value).

[L3]

Triangle inequality: ∣u+v∣≤∣u∣+∣v∣ (The triangle inequality).

[L4]

Order and field arithmetic in R: adding two strict inequalities (Order is preserved by adding a constant and by adding inequalities); for t>0, u<v is equivalent to ut<vt, and 0≤u≤v with 0≤s≤t gives us≤vt (Sign rules for products and monotonicity of multiplication); positive elements have positive inverses and 0<a<b gives 0<1/b<1/a (Inverses of positives are positive, and reciprocation reverses order); 0<1 (The multiplicative identity is positive), so 2>0 and t/2>0 for t>0; inverses and the field identities (Field); trichotomy and totality, so of finitely many positive reals the smallest is positive (Ordered field).

[L5]

Local boundedness: there are a real δ0>0 and a real K≥0 with ∣f(x)∣≤K for every x∈A satisfying 0<∣x−c∣<δ0 (If f has a finite limit at c then f is bounded on some punctured neighbourhood of c).

[L6]

Sign preservation: if M≠0 there is a real δs>0 with ∣g(x)∣>∣M∣/2>0 for every x∈A satisfying 0<∣x−c∣<δs, and c is a limit point of A0 (If lim⁡x→cf(x)=L≠0 then ∣f∣>∣L∣/2 on a punctured neighbourhood of c; in particular if L>0 then f>L/2>0 there).

[L7]

Restriction: if B⊆A has c as a limit point and lim⁡x→cf(x)=L, then lim⁡x→cf∣B(x)=L (claim 2 of 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).

Proof

technique · direct
1.1

Sum. Let ε>0 be an arbitrary real. By [L1] fix reals δ1,δ2>0 with ∣f(x)−L∣<ε/2 for every x∈A satisfying 0<∣x−c∣<δ1 and ∣g(x)−M∣<ε/2 for every x∈A satisfying 0<∣x−c∣<δ2, and let δ be the smaller of the two, so δ>0. For x∈A with 0<∣x−c∣<δ we get ∣(f+g)(x)−(L+M)∣=∣(f(x)−L)+(g(x)−M)∣≤∣f(x)−L∣+∣g(x)−M∣<ε. As ε was arbitrary, the limit of f+g at c exists and equals L+M: claim 1.

L1L2L3L4choose
1.2

Scalar multiple. If α=0 then αf is the constant function 0 and αL=0, so ∣(αf)(x)−αL∣=0<ε for every x and every ε>0, any δ serving. If α≠0 then ∣α∣>0; given a real ε>0, [L1] supplies δ>0 with ∣f(x)−L∣<ε/∣α∣ on A∩Nδ∗(c), and there ∣(αf)(x)−αL∣=∣α∣ ∣f(x)−L∣<ε. So the limit of αf at c exists and equals αL: claim 2.

L1L2L4L8choose
1.3

A working bound for f near c. By [L5] fix a real δ0>0 and a real K≥0 with ∣f(x)∣≤K for every x∈A satisfying 0<∣x−c∣<δ0, and put K′:=K+1, so K′>0 and ∣f(x)∣≤K′ for all those x.

L4L5choose
1.4

The denominator near c. Assume M≠0. By [L6] fix a real δs>0 with ∣g(x)∣>∣M∣/2>0 for every x∈A satisfying 0<∣x−c∣<δs; every such x has g(x)≠0, hence lies in A0, and c is a limit point of A0.

L2L4L6
2.1

Product. Let ε>0 be an arbitrary real. By [L1] fix reals δ1,δ2>0 with ∣g(x)−M∣<ε/(2K′) on A∩Nδ1∗(c) and ∣f(x)−L∣<ε/(2(∣M∣+1)) on A∩Nδ2∗(c), and let δ be the smallest of δ0,δ1,δ2, which is positive. For x∈A with 0<∣x−c∣<δ, ∣f(x)g(x)−LM∣=∣f(x)(g(x)−M)+M(f(x)−L)∣≤∣f(x)∣ ∣g(x)−M∣+∣M∣ ∣f(x)−L∣≤K′ ∣g(x)−M∣+(∣M∣+1) ∣f(x)−L∣<ε/2+ε/2=ε. As ε was arbitrary, the limit of fg at c exists and equals LM: claim 3.

step 1.3L1L2L3L4L8choose
2.2

Reciprocal. Assume M≠0 and let ε>0 be an arbitrary real. By [L1] fix a real δ3>0 with ∣g(x)−M∣<ε∣M∣2/2 on A∩Nδ3∗(c), and let δ be the smaller of δs and δ3. For x∈A0 with 0<∣x−c∣<δ we have ∣g(x)∣>∣M∣/2>0, hence ∣g(x)∣ ∣M∣>∣M∣2/2>0 and so 1/(∣g(x)∣ ∣M∣)<2/∣M∣2; therefore ∣1/g(x)−1/M∣=∣M−g(x)∣/(∣g(x)∣ ∣M∣)<(ε∣M∣2/2)⋅(2/∣M∣2)=ε. As ε was arbitrary, the limit of (1/g)∣A0 at c exists and equals 1/M.

step 1.4L1L2L4L8choose
2.3

The numerator on the smaller domain. Assume M≠0. Since A0⊆A and c is a limit point of A0 by step 1.4, [L7] gives that the limit of f∣A0 at c exists and equals L.

step 1.4L7
3.1

Quotient. Assume M≠0. On the domain A0, which has c as a limit point, the two functions f∣A0 and (1/g)∣A0 have limits L and 1/M at c by steps 2.3 and 2.2, and their product is (f/g)∣A0 by the field identities; so claim 3, applied on the domain A0, gives that the limit of (f/g)∣A0 at c exists and equals L⋅(1/M)=L/M.

step 2.1step 2.2step 2.3L2L4
4.1

Claims 1 to 4 are proved, each directly from the ε-δ definition and none of them through a sequence.

step 1.1step 1.2step 2.1step 3.1∎

Remarks

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

If f≤g on a punctured neighbourhood of c then lim⁡f≤lim⁡g, non-strictly

Statement

Let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), let f,g:A→R and suppose both limits at c exist (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Suppose further that there is a real η>0 with

f(x)≤g(x)for every x∈A with 0<∣x−c∣<η.

Then

lim⁡x→cf(x)  ≤  lim⁡x→cg(x).

The conclusion is non-strict even when the hypothesis is strict. Replacing ≤ by < on both sides gives a false statement, refuted by FALSE: f<g near c implies lim⁡f<lim⁡g: strictness is destroyed in the limit, and no hypothesis short of a uniform gap restores it.

Only the values near c matter, by 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: the hypothesis is imposed on a punctured neighbourhood of c and on nothing else, and it says nothing about f(c) and g(c), which the definition ignores in any case.

Facts & Assumptions

Given: A set A⊆R, a limit point c of A, functions f,g:A→R, reals L,M with lim⁡x→cf(x)=L and lim⁡x→cg(x)=M, and a real η>0 with f(x)≤g(x) for every x∈A satisfying 0<∣x−c∣<η (The ε-δ limit lim⁡x→cf(x)=L of f:A→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∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε, and likewise for g and M (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]
[L3]

Absolute value: for t>0, ∣u∣<t is equivalent to −t<u<t (Basic properties of the absolute value).

[L4]

Order arithmetic in R: the order is total, so the negation of u≤v is v<u; trichotomy, so u<v and v≤u cannot both hold; adding a constant to an inequality and adding two inequalities (Order is preserved by adding a constant and by adding inequalities); 0<1 (The multiplicative identity is positive), so 2>0, 2−1>0 (Inverses of positives are positive, and reciprocation reverses order) and t/2>0 for t>0 (Sign rules for products and monotonicity of multiplication), with (t/2)+(t/2)=t; and of finitely many positive reals the smallest is positive (Ordered field).

Proof

technique · contradiction
1.1

Suppose, for contradiction, that L≤M fails; the order being total, this means M<L.

assume-contra
2.1

Then L−M>0, so ε:=(L−M)/2>0, and L−ε=(L+M)/2=M+ε.

step 1.1L4
3.1

By [L1] fix reals δ1,δ2>0 such that every x∈A with 0<∣x−c∣<δ1 has ∣f(x)−L∣<ε and every x∈A with 0<∣x−c∣<δ2 has ∣g(x)−M∣<ε; let δ be the smallest of δ1, δ2 and η, so δ>0.

step 2.1L1L4choose
4.1

Since c is a limit point of A, fix x∈A with 0<∣x−c∣<δ.

step 3.1L2choose
5.1

That x satisfies 0<∣x−c∣<δ1 and 0<∣x−c∣<δ2, so ∣f(x)−L∣<ε gives f(x)>L−ε and ∣g(x)−M∣<ε gives g(x)<M+ε; since L−ε=M+ε, this yields g(x)<f(x).

step 3.1step 4.1L3L4
6.1

But that same x satisfies 0<∣x−c∣<η, so the hypothesis gives f(x)≤g(x), which together with g(x)<f(x) contradicts trichotomy.

step 3.1step 5.1L4
7.1

The assumption that L≤M fails is therefore untenable, and lim⁡x→cf(x)=L≤M=lim⁡x→cg(x).

step 6.1L4discharge-contradiction∎

Remarks

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

If f≤g≤h near c and f and h have the same limit at c, then so does g

Statement

Let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R) and let f,g,h:A→R. Suppose there is a real η>0 with

f(x)≤g(x)≤h(x)for every x∈A with 0<∣x−c∣<η,

and suppose the limits of f and of h at c exist and are equal, say lim⁡x→cf(x)=lim⁡x→ch(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Then the limit of g at c exists, and

lim⁡x→cg(x)  =  lim⁡x→cf(x)  =  lim⁡x→ch(x)  =  L.

This is the one result on this page that produces a limit rather than computing one. No hypothesis whatever is placed on g beyond the two inequalities: g may be wildly irregular, as x↦x ψ(1/x) on the companion page is, and the theorem still delivers its limit at c.

The proof is a direct ε-δ argument and uses no choice principle.

Facts & Assumptions

Given: A set A⊆R, a limit point c of A, functions f,g,h:A→R, a real η>0 with f(x)≤g(x)≤h(x) for every x∈A satisfying 0<∣x−c∣<η, and a real L with lim⁡x→cf(x)=L and lim⁡x→ch(x)=L (The ε-δ limit lim⁡x→cf(x)=L of f:A→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∈A with 0<∣x−c∣<δ satisfies ∣f(x)−L∣<ε, and likewise for h (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Absolute value: for t>0, ∣u∣<t is equivalent to −t<u<t (Basic properties of the absolute value).

[L3]

Order arithmetic in R: the order is transitive, and mixed chains compose, so u<v≤w gives u<w and u≤v<w gives u<w; adding a constant to an inequality (Order is preserved by adding a constant and by adding inequalities); of finitely many positive reals the smallest is positive, the order being total (Ordered field). Order is preserved by adding a constant and by adding inequalities states its moves in their STRICT forms only; the non-strict forms used below follow by adjoining the equality case, in which the two sides coincide, the order being total (Ordered field).

[L4]

Neighbourhoods: Nδ∗(c)={ y:0<∣y−c∣<δ }, and a smaller radius gives a smaller punctured neighbourhood (The ε-neighbourhood and the punctured ε-neighbourhood of a point of R).

Proof

technique · direct
1.1

Let ε>0 be an arbitrary real. By [L1] fix reals δ1,δ2>0 such that every x∈A with 0<∣x−c∣<δ1 satisfies ∣f(x)−L∣<ε and every x∈A with 0<∣x−c∣<δ2 satisfies ∣h(x)−L∣<ε; let δ be the smallest of δ1, δ2 and η, so δ>0.

L1L3L4choose
2.1

Let x∈A with 0<∣x−c∣<δ. Then 0<∣x−c∣<δ1 gives L−ε<f(x), and 0<∣x−c∣<δ2 gives h(x)<L+ε, while 0<∣x−c∣<η gives f(x)≤g(x)≤h(x).

step 1.1L2L3L4
3.1

Chaining those four inequalities, L−ε<f(x)≤g(x)≤h(x)<L+ε, hence L−ε<g(x)<L+ε, that is −ε<g(x)−L<ε, that is ∣g(x)−L∣<ε.

step 2.1L2L3
4.1

So for every real ε>0 a real δ>0 has been produced with ∣g(x)−L∣<ε for every x∈A satisfying 0<∣x−c∣<δ: the limit of g at c exists and equals L.

step 3.1L1∎

Remarks

  • Where the three hypotheses are spent. The inequality f≤g is used only for the lower estimate and g≤h only for the upper one; the equality of the two outer limits is what makes the two estimates close on the same number L. Drop it and the argument gives only lim⁡f≤lim inf⁡-style information, which this page does not develop.

  • The order hypothesis is local. It is imposed only on A∩Nη∗(c), so the theorem is insensitive to the behaviour of the three functions far from c, and to their values at c; that is 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 in action.

  • Typical use. To prove that a bounded oscillating factor is killed by a factor tending to 0: if ∣u(x)∣≤B near c then −B∣x−c∣≤(x−c)u(x)≤B∣x−c∣ near c, and both outer functions tend to 0. That is exactly how x ψ(1/x)→0 is proved on the companion page.

  • The sequential analogue is The squeeze theorem.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

If c is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree

Statement

Let A⊆R, let f:A→R and let c∈R be a limit point of both A−=A∩(−∞,c) and A+=A∩(c,∞) (Limit point, isolated point, adherent point, derived set, and dense subset of R, Intervals of R: the nine order-convex forms, nondegeneracy, and length), so that both one-sided limits at c are well posed (The left and right limits of f at c, as limits of the restrictions of f to A∩(−∞,c) and A∩(c,∞)). Then c is a limit point of A, and for every L∈R:

lim⁡x→cf(x)=L⟺lim⁡x→c−f(x)=L  and  lim⁡x→c+f(x)=L

(The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Consequently the limit of f at c exists if and only if both one-sided limits exist and are equal, and in that case

lim⁡x→cf(x)  =  lim⁡x→c−f(x)  =  lim⁡x→c+f(x).

The hypothesis on both sides is what makes the statement an equivalence. If c is a limit point of only one of the two sets — as 1 is for {0}∪[1,2] — then the one-sided limit on that side and the two-sided limit are the same condition, and the symbol on the other side is not defined at all (The left and right limits of f at c, as limits of the restrictions of f to A∩(−∞,c) and A∩(c,∞)).

Facts & Assumptions

[L1]

The limit condition (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A): lim⁡x→ch(x)=L means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−L∣<ε.

[L2]

Limit point: c is a limit point of S when for every real δ>0 there is x∈S 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]

Intervals: A−={ x∈A:x<c } and A+={ x∈A:x>c } (Intervals of R: the nine order-convex forms, nondegeneracy, and length).

[L4]

Absolute value and order: ∣x−c∣=0 exactly when x=c; the order is total, so every x≠c satisfies x<c or x>c; and 0<∣x−c∣<δ is equivalent to c−δ<x<c for x<c and to c<x<c+δ for x>c (Basic properties of the absolute value, Ordered field). Of two positive reals the smaller is positive.

[L5]

Restriction: if B⊆A has c as a limit point and lim⁡x→cf(x)=L, then lim⁡x→cf∣B(x)=L (claim 2 of 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).

[L6]

One-sided limits are by definition the limits of the restrictions f∣A− and f∣A+ at c (The left and right limits of f at c, as limits of the restrictions of f to A∩(−∞,c) and A∩(c,∞)).

[L7]

At a limit point of its domain a function has at most one limit (At a limit point of the domain a function has at most one limit); applied to f∣A− and to f∣A+ it makes each one-sided limit a single real, and applied to f it does the same for the two-sided limit.

Proof

technique · direct
1.1

c is a limit point of A: it is one of A+ by hypothesis, and A+⊆A, so every point of A+ found in a punctured neighbourhood of c is a point of A there.

L2L3
1.2

For x∈A the condition 0<∣x−c∣ says exactly x≠c, and then x<c or x>c, that is x∈A− or x∈A+; moreover for x∈A− the condition 0<∣x−c∣<δ reads c−δ<x<c and for x∈A+ it reads c<x<c+δ.

L3L4
2.1

Suppose lim⁡x→cf(x)=L. Both A− and A+ are subsets of A having c as a limit point, so [L5] gives lim⁡x→cf∣A−(x)=L and lim⁡x→cf∣A+(x)=L, which by [L6] is exactly lim⁡x→c−f(x)=L and lim⁡x→c+f(x)=L.

step 1.1step 1.2L5L6
2.2

Suppose conversely that both one-sided limits equal L, and let ε>0 be an arbitrary real. By [L6] and [L1] fix reals δ1,δ2>0 such that every x∈A− with 0<∣x−c∣<δ1 and every x∈A+ with 0<∣x−c∣<δ2 satisfies ∣f(x)−L∣<ε; let δ be the smaller of the two. Every x∈A with 0<∣x−c∣<δ lies in A− or in A+ by step 1.2, and in either case ∣f(x)−L∣<ε. As ε was arbitrary, lim⁡x→cf(x)=L.

step 1.2L1L4L6choose
3.1

The displayed equivalence is steps 2.1 and 2.2. For the consequence: if the limit of f at c exists, say with value L, then step 2.1 gives that both one-sided limits exist with the same value L, so they agree; and if both one-sided limits exist and are equal, to the common value L, then step 2.2 gives that the limit of f at c exists and equals L. Each of the three symbols denotes a single real by [L7], so the three are equal.

step 2.1step 2.2L7∎

Remarks

  • The two directions are not symmetric in difficulty. From the two-sided limit to the one-sided ones is pure restriction, 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; the converse has to glue two estimates, and the gluing is legitimate precisely because every point of A other than c lies strictly on one side of c, which is the totality of the order.

  • The typical failure is a function whose two one-sided limits exist and differ: the sign function at 0, on the companion page. Then the two-sided limit cannot exist, since by step 2.1 it would force both one-sided values to equal it.

  • A function may also have no two-sided limit for a different reason, namely that a one-sided limit fails to exist rather than that the two disagree. The theorem covers that case too, since its right-hand side asserts the existence of both one-sided values, so its failure on one side alone already blocks the two-sided limit. The companion page exhibits both patterns.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c

Statement

Let A,B⊆R, let g:A→R with g(A)⊆B, and let f:B→R, so that the composite f∘g:A→R is defined. Let c be a limit point of A and L a limit point of B (Limit point, isolated point, adherent point, derived set, and dense subset of R), and suppose the limits

lim⁡x→cg(x)=Landlim⁡y→Lf(y)=M

both exist, with the stated values (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Suppose in addition that at least one of the following holds:

  • (i) L∈B and f(L)=M;
  • (ii) there is a real η>0 with g(x)≠L for every x∈A satisfying 0<∣x−c∣<η.

Then the limit of f∘g at c exists, and

lim⁡x→cf(g(x))  =  lim⁡y→Lf(y)  =  M.

At least one extra hypothesis is necessary. With both omitted the statement is false, and FALSE: lim⁡x→cf(g(x))=M whenever lim⁡x→cg=L and lim⁡y→Lf=M refutes it with a two-line witness in which (i) fails because f(L)≠M and (ii) fails because g is constantly equal to L.

Why an extra hypothesis is needed at all. The inner limit controls g(x) only up to ∣g(x)−L∣<ρ; it does not prevent g(x) from equalling L. But The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A says nothing about f at the point L, so the outer estimate is unavailable exactly at the values g(x)=L. Hypothesis (i) supplies the missing value directly; hypothesis (ii) excludes those values.

Facts & Assumptions

Given: Sets A,B⊆R, functions g:A→R with g(A)⊆B and f:B→R, a limit point c of A, a limit point L of B, and reals with lim⁡x→cg(x)=L and lim⁡y→Lf(y)=M; and the assumption that (i) or (ii) of the statement holds (The ε-δ limit lim⁡x→cf(x)=L of f:A→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: lim⁡x→ch(x)=P means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−P∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Absolute value: ∣u∣≥0, and ∣u∣=0 if and only if u=0 (Basic properties of the absolute value).

[L3]

Order arithmetic: of two positive reals the smaller is positive, the order being total; and trichotomy (Ordered field).

Proof

technique · cases
1.1

Let ε>0 be an arbitrary real. By [L1] applied to f at L, fix a real ρ>0 such that every y∈B with 0<∣y−L∣<ρ satisfies ∣f(y)−M∣<ε; then by [L1] applied to g at c, with ρ in the role of the tolerance, fix a real δ1>0 such that every x∈A with 0<∣x−c∣<δ1 satisfies ∣g(x)−L∣<ρ.

L1choose
2.1

Case (i): assume L∈B and f(L)=M, and put δ:=δ1>0. Let x∈A with 0<∣x−c∣<δ and set y:=g(x), an element of B since g(A)⊆B; then ∣y−L∣<ρ. If y=L then ∣f(y)−M∣=∣f(L)−M∣=∣0∣=0<ε; and if y≠L then 0<∣y−L∣<ρ, so ∣f(y)−M∣<ε. In both events ∣(f∘g)(x)−M∣<ε.

step 1.1assume-case valueL1L2L3
2.2

Case (ii): assume there is a real η>0 with g(x)≠L for every x∈A satisfying 0<∣x−c∣<η, and let δ be the smaller of δ1 and η, so δ>0. Let x∈A with 0<∣x−c∣<δ and set y:=g(x)∈B; then y≠L, so ∣y−L∣>0, and ∣y−L∣<ρ, so 0<∣y−L∣<ρ and ∣(f∘g)(x)−M∣=∣f(y)−M∣<ε.

step 1.1assume-case avoidL1L2L3
3.1

By hypothesis at least one of (i) and (ii) holds, so in either case a real δ>0 has been produced with ∣(f∘g)(x)−M∣<ε for every x∈A satisfying 0<∣x−c∣<δ; since ε>0 was arbitrary and c is a limit point of A, the limit of f∘g at c exists and equals M.

step 2.1step 2.2L1L4cases-exhaustive∎

Remarks

  • The hypothesis that L is a limit point of B is what makes lim⁡y→Lf(y) meaningful at all (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A); it is not an extra assumption of convenience. Note that it does not follow from lim⁡x→cg(x)=L: a constant g has that limit while B may be a set for which L is isolated.

  • Hypothesis (i) is the continuity hypothesis in disguise. Saying L∈B and f(L)=M=lim⁡y→Lf(y) is exactly saying that f is continuous at L in the sense the next page of this track will define; that is the form in which this theorem is usually quoted, and it is why textbook statements of "the limit of a composition" almost always assume continuity of the outer function.

  • Hypothesis (ii) is the one that survives without continuity, and it is the hypothesis under which substitutions such as y=1/x are legitimate: there the inner function omits the critical value on a punctured neighbourhood for a structural reason, not by assumption on f.

  • The two hypotheses are genuinely different, neither implying the other. The companion page exhibits a pair satisfying neither, and the same pair with the inner function replaced by the identity, which satisfies (ii) but not (i).

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

Integer part: for every real x there is exactly one integer m with m≤x<m+1

Statement

Identify Z with its canonical copy inside R, along the embeddings N→Z→Q→R (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, The integers as equivalence classes of pairs of naturals). Then for every real x there is exactly one integer m with

m  ≤  x  <  m+1.

It is written ⌊x⌋ and called the integer part, or floor, of x.

Two independent ingredients are needed and neither may be dropped. Existence is the Archimedean property (Every complete ordered field is Archimedean) together with the well-ordering of N (The well-ordering principle): the first says that x is caught between two integers at all, the second picks the least integer above x. Uniqueness is the discreteness of Z: no integer lies strictly between m and m+1.

This lemma is stated once here and reused. It is what turns "the nearest integer to x" from a picture into an object, and the companion page's oscillator ψ(x)=inf⁡n∈Z∣x−n∣ is computed from it in one line.

Facts & Assumptions

Given: A real x. Naturals, integers and rationals are identified with their canonical copies in R along N→Z→Q→R.

[L1]

The embeddings N→Z→Q→R are injective and preserve 0, 1, addition, multiplication and order (The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, The integers as equivalence classes of pairs of naturals); Z is a totally ordered commutative ring (The integers form a totally ordered ring, The integers form a commutative ring); every integer ≥0 is the image of a unique natural, that map being injective and order preserving (The naturals embed in the integers); and a natural j≠0 satisfies j≥1 (Discreteness: σ(n) is the immediate successor, The natural numbers N (von Neumann)).

[L2]

The image of a natural n≥1 under the composite N→R is the canonical natural n⋅1R of Canonical naturals are positive and strictly increasing. Indeed that composite preserves 1 and addition by [L1], while n⋅1R is defined by 1⋅1R=1R and (n+1)⋅1R=n⋅1R+1R, so the two agree at 1 and satisfy the same recursion; induction on n (The principle of mathematical induction) gives the identification.

[L3]

Archimedean property: for every real t there is a natural n≥1 with t<n⋅1R (Every complete ordered field is Archimedean, Complete ordered field (least-upper-bound property)).

[L4]

Well-ordering principle: every nonempty subset of N has a least element (The well-ordering principle).

[L5]

Order arithmetic in R: the order is total, so the negation of t<u is u≤t; trichotomy, so t<u and u≤t cannot both hold; translation invariance (Order is preserved by adding a constant and by adding inequalities); t≤∣t∣ and −t≤∣t∣ (Basic properties of the absolute value); and transitivity (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · constructive
1.1

Apply [L3] to the real ∣x∣: fix a natural n≥1 with ∣x∣<n. Since x≤∣x∣ and −x≤∣x∣, this gives −n<x<n.

L2L3L5choose
2.1

Put S:={ k∈N : x<k−n }, where k−n is formed in Z and read in R through [L1]. It is a subset of N, and it is nonempty: the natural 2n satisfies 2n−n=n>x by step 1.1, so 2n∈S.

step 1.1L1L2construct
3.1

By the well-ordering principle [L4] let k0 be the least element of S.

step 2.1L4choose
4.1

The index k0 is not 0: for k=0 the defining condition reads x<0−n=−n, which trichotomy excludes since −n<x by step 1.1. Hence k0≠0, so k0≥1 by [L1], and k0−1 is again a natural number.

step 1.1step 3.1L1L5
5.1

Set m:=(k0−1)−n, an integer. Since k0−1<k0 and k0 is the least element of S, the natural k0−1 does not lie in S, that is, x<(k0−1)−n fails; the order being total, m=(k0−1)−n≤x.

step 3.1step 4.1L1L5construct
6.1

On the other hand k0∈S gives x<k0−n=((k0−1)−n)+1=m+1. So m≤x<m+1, and existence is proved.

step 3.1step 5.1L1L5
7.1

Uniqueness: suppose an integer m′ also satisfies m′≤x<m′+1 and m′≠m. The order of Z being total, one of m<m′ and m′<m holds, and the two cases are the same with the roles of m and m′ exchanged; so assume m<m′. Then m′−m is an integer >0, hence by [L1] the image of a natural j≠0, so j≥1 and m′−m≥1, that is m+1≤m′. But then x<m+1≤m′≤x, which trichotomy forbids. Hence m′=m.

step 6.1L1L5
8.1

Therefore exactly one integer m satisfies m≤x<m+1, and we write m=⌊x⌋.

step 6.1step 7.1discharge-construct∎

Remarks

  • What the two halves of the proof really use. Step 1.1 is the only use of the Archimedean property, and it is indispensable: in a non-Archimedean ordered field (Not every ordered field is Archimedean) an element larger than every canonical natural has no integer part at all, since the set S of step 2.1 would be empty. Step 3.1 is the only use of the well-ordering principle, and it is what makes the construction canonical: no choice is made anywhere, and ⌊x⌋ is a function of x.

  • Immediate consequences, used later. From m≤x<m+1 one reads off 0≤x−m<1 and 0<(m+1)−x≤1; and ⌊x⌋=x exactly when x is an integer, since an integer m satisfies m≤m<m+1 and uniqueness does the rest. The translation identity ⌊x+p⌋=⌊x⌋+p for an integer p follows the same way: adding p to m≤x<m+1 gives m+p≤x+p<(m+p)+1, and uniqueness identifies m+p as the integer part of x+p.

  • The ceiling is not defined here and is not needed on this page; it would be the least integer ≥x, obtained from the same set S without the shift by one.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

The sequence-to-ε direction of the Heine criterion uses countable choice for R, and where this library records that cost

What this page spends, and where

Heine criterion: lim⁡x→cf(x)=L iff f(xk)→L for every sequence in A∖{c} converging to c is an equivalence, and its two directions do not cost the same.

  • From the ε-δ limit to sequences — if lim⁡x→cf(x)=L then f(xk)→L for every sequence in A∖{c} tending to c — is proved in ZF. The sequence is handed to the proof; nothing is selected. This is steps 1.1, 2.1 and 3.1 of that theorem.

  • From sequences to the ε-δ limit is proved there using the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)), invoked exactly once, at step 3.2. The proof assumes the limit fails, obtains for each k∈N a nonempty set Xk={ x∈A:0<∣x−c∣<1/(k+1) and ∣f(x)−L∣≥ε0 }, and needs a single point from each of those countably many sets at once.

Why no canonical selection is available. The sets Xk are cut out by an inequality involving f, about which the theorem assumes nothing. There is therefore no rule in this library that names an element of Xk uniformly in k: they are subsets of R, which carries no well-ordering that ZF provides, and the sets need not be intervals, need not be closed, and need not meet Q. That is precisely the situation The Axiom of Countable Choice (ACω) exists for.

The same cost, recorded twice

The identical pattern occurs in the prerequisite page: in A point lies in the closure of A⊆R iff some sequence in A converges to it, so a subset of R is closed iff it is sequentially closed the right-to-left direction is choice free, while producing a sequence in A converging to a point of A‾ requires selecting one point of A from each of the sets N1/(k+1)(x)∩A, and that item invokes ACω explicitly for it. Both items name the step where the axiom is used, so a reader working in ZF alone can see exactly which half of each equivalence survives.

What this library claims, and what it does not

  • Claimed: the direction from ε-δ to sequences is a theorem of ZF; the converse as proved here uses ACω; and the use is isolated to one step, so nothing else on this page inherits it.

  • Not claimed: that the converse requires ACω. This library proves no independence result and contains neither forcing nor permutation models, so it is in no position to assert that some cleverer ZF proof does not exist. The systematic study of which such criteria need which fragment of choice is a subject in its own right; Herrlich's Axiom of Choice is the standard reference, and it is cited here as literature, not used.

  • A warning against a tempting slogan. It is not the case that sequential criteria in analysis always need choice. Sierpiński proved, in ZF, that a function R→R which is sequentially continuous at every point is continuous. The everywhere-statement and the pointwise-statement behave differently, and the cost recorded above is a statement about the pointwise criterion as proved here, nothing more.

The consequence for how this page is organised

Because the criterion carries a choice cost on one side, this page does not route its main results through it. The algebra of limits (Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero), order preservation (If f≤g on a punctured neighbourhood of c then lim⁡f≤lim⁡g, non-strictly), the squeeze theorem (If f≤g≤h near c and f and h have the same limit at c, then so does g) and composition (Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c) are all proved directly from ε and δ, and are therefore theorems of ZF. The sequential machinery is used only where it earns its place: in the criterion itself, and in A function has no limit at c as soon as two sequences in A∖{c} tending to c give different limits of the values, which needs only the choice-free direction and is the tool by which the companion page shows that various limits fail to exist.

That organisation is a deliberate choice of proofs, not a mathematical necessity: each of those four results could be deduced from the criterion, at the price of importing ACω into statements that do not need it.

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist

Statement

False claim: if A⊆R, if f:A→R, if c∈A is a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R) and if the limit of f at c exists (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A), then

lim⁡x→cf(x)=f(c).

Both sides of the asserted equation are defined under the stated hypotheses: the left because the limit is assumed to exist and is single valued (At a limit point of the domain a function has at most one limit), the right because c∈A. The claim is that they always agree, and that is false.

Why it is tempting. The condition ∣f(x)−L∣<ε is imposed on points x arbitrarily close to c, and it feels as though x=c were the limiting case of that. It is not: The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A quantifies over 0<∣x−c∣<δ, and the strict inequality on the left removes x=c from the quantifier entirely. Changing the value of f at the single point c therefore changes nothing on the left-hand side and everything on the right.

What is true. The equation above is not a theorem but a condition, and it is the condition the next page of this track takes as the definition of continuity at c. This library states it as a hypothesis and never as a consequence; hypothesis (i) of Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c is exactly this condition for the outer function.

Facts & Assumptions

Given: The set A:=R, the point c:=0, and the function f:R→R defined by f(x):=0 for x≠0 and f(0):=1.

[L1]

The limit condition: lim⁡x→ch(x)=L means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−L∣<ε (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

[L2]

Limit point: c is a limit point of S when every punctured neighbourhood Nε∗(c) meets S; and punctured neighbourhoods in R are never empty (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]

Absolute value: ∣0∣=0, and ∣u∣≥0 (Basic properties of the absolute value).

[L4]

Order in R: trichotomy, so every real either equals 0 or does not, and never both; and 0<1, so 1≠0 (The multiplicative identity is positive, Ordered field).

Refutation

technique · direct
1.1

The point c=0 lies in A=R and is a limit point of R: for every real ε>0 the punctured neighbourhood Nε∗(0) is nonempty and is contained in R, so it meets R.

L2
1.2

f is a well-defined function on R, since by trichotomy every real either equals 0 or does not, exclusively; and the reals 0 and 1 are distinct.

L4
2.1

The limit of f at 0 exists and equals 0: given an arbitrary real ε>0, take δ:=1>0; every x∈R with 0<∣x−0∣<1 has ∣x∣≠0, hence x≠0, hence f(x)=0 and ∣f(x)−0∣=∣0∣=0<ε.

step 1.1step 1.2L1L3L4
3.1

Yet f(0)=1, and 1≠0=lim⁡x→0f(x). So at the point c=0 of the domain, which is a limit point of the domain, the limit exists and differs from the value: the claim is false.

step 1.2step 2.1L4∎

Remarks

  • The witness is the smallest possible one. It differs from a constant function at exactly one point, and the limit cannot see that point. Any function agreeing with a constant off c and taking a different value at c would serve equally well; the companion page works this witness out in full, computes its one-sided limits, and shows that redefining the single value repairs the equality.

  • Where the false claim does hold. Under the extra hypothesis that lim⁡x→cf(x)=f(c) — which is what continuity at c will mean — it holds trivially, and that is the only sense in which it is ever true. It is emphatically not a consequence of the limit existing.

  • The consequence for composition. Because f(c) is invisible to the limit, substituting an inner function that takes the value c is not licensed by the limits alone; that is the content of FALSE: lim⁡x→cf(g(x))=M whenever lim⁡x→cg=L and lim⁡y→Lf=M, whose witness is built from this one.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: lim⁡x→cf(g(x))=M whenever lim⁡x→cg=L and lim⁡y→Lf=M

Statement

False claim: let A,B⊆R, let g:A→R with g(A)⊆B and f:B→R, let c be a limit point of A and L a limit point of B. If

lim⁡x→cg(x)=Landlim⁡y→Lf(y)=M,

then the limit of f∘g at c exists and lim⁡x→cf(g(x))=M (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A).

This is the statement of Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c with both of its extra hypotheses removed, and it is false. It is refuted below by a pair in which g is constant and f has a removable defect at the value of that constant.

Where the naive argument breaks. The inner limit gives ∣g(x)−L∣<ρ for x near c; the outer limit gives ∣f(y)−M∣<ε for y∈B with 0<∣y−L∣<ρ. To combine them at y=g(x) one needs ∣g(x)−L∣>0, and nothing in the hypotheses supplies that. Where g(x)=L, the only information available about f is its value f(L), and The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A says nothing whatever about that value (FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist). The two hypotheses of Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c are exactly the two ways of closing that gap.

Facts & Assumptions

Given: The sets A:=R and B:=R; the point c:=0; the function f:R→R of FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist, namely f(y):=0 for y≠0 and f(0):=1; and the constant function g:R→R, g(x):=0 for every x.

[L1]

The limit condition (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A): lim⁡x→ch(x)=P means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−P∣<ε.

[L3]

Absolute value: ∣0∣=0 (Basic properties of the absolute value).

[L4]

Order in R: trichotomy, and 0<1, so 1≠0 (The multiplicative identity is positive, Ordered field).

[L5]

The function f above satisfies f(0)=1 and has limit 0 at 0: for every real ε>0 the radius δ=1 works, since 0<∣y−0∣<1 forces y≠0 and then ∣f(y)−0∣=0<ε; this is the computation carried out in FALSE: lim⁡x→cf(x)=f(c) whenever both sides exist.

Refutation

technique · direct
1.1

The point 0 is a limit point of R, and g(R)={0}⊆R=B, so f∘g is a function on R.

L2
1.2

By [L5], lim⁡y→0f(y)=0; so the outer hypothesis holds with L=0 and M=0.

L5
1.3

The reals 0 and 1 are distinct.

L4
2.1

The inner hypothesis holds with L=0: for the constant function g and any real ε>0, every δ>0 works, since ∣g(x)−0∣=∣0∣=0<ε for every x. So lim⁡x→0g(x)=0.

step 1.1L1L3
3.1

But f∘g is the constant function 1: for every x∈R, g(x)=0 and hence f(g(x))=f(0)=1. Therefore, by the same computation as in step 2.1, the limit of f∘g at 0 exists and equals 1.

step 2.1L1L3L5
3.2

Both extra hypotheses of Composition of limits holds under either hypothesis: f is defined at L with value M, or g avoids L on a punctured neighbourhood of c fail for this pair: hypothesis (i) fails because L=0 lies in B=R while f(L)=f(0)=1≠0=M; and hypothesis (ii) fails because g(x)=0=L for every x, so no punctured neighbourhood of 0 avoids the value L.

step 2.1L5L6
4.1

So lim⁡x→0g(x)=0=L and lim⁡y→0f(y)=0=M, while lim⁡x→0f(g(x))=1≠0=M: the claim is false, and step 3.2 identifies exactly which hypotheses of the true theorem are missing.

step 1.2step 1.3step 3.1step 3.2∎

Remarks

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: a function has at most one limit at every point of its domain, isolated points included

Statement

False claim: for every A⊆R, every f:A→R and every c∈A, at most one real L satisfies

(∀ε>0) (∃δ>0) (∀x∈A) [ 0<∣x−c∣<δ ⟹ ∣f(x)−L∣<ε ].(∗)

Read the claim carefully: it is about the raw formula (∗), extended to an arbitrary point c of the domain. It is not a claim about The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A. That definition imposes (∗) only when c is a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), and there at most one L does satisfy it — that is exactly At a limit point of the domain a function has at most one limit, which is true and proved. The false claim is what one gets by deleting the limit-point requirement.

At an isolated point of A the symbol lim⁡x→cf(x) is undefined in this library, and the refutation below is the reason. If c∈A is not a limit point of A then some punctured neighbourhood of c misses A entirely (Limit point, isolated point, adherent point, derived set, and dense subset of R); the implication inside (∗) then has no instances at all for that δ, so it holds vacuously, and it holds for every real L at once. A formula satisfied by every real determines nothing, so no notation is introduced for it.

Facts & Assumptions

Given: The set A:={0}∪[1,2] (Intervals of R: the nine order-convex forms, nondegeneracy, and length), the constant function f:A→R with f(x):=0 for every x∈A, and the point c:=0∈A.

[L1]

The ε-δ formula (∗) above, and the fact that The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A imposes it only at a limit point of the domain.

[L2]

Limit point and isolated point: c is a limit point of S when Nε∗(c)∩S≠∅ for every real ε>0, and c∈S is an isolated point of S when Nε(c)∩S={c} for some real ε>0; for c∈S these are exact opposites (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]

Neighbourhoods: N1(0)={ y:∣y∣<1 } and N1∗(0)={ y:0<∣y∣<1 } (The ε-neighbourhood and the punctured ε-neighbourhood of a point of R).

[L4]

Absolute value and order: ∣u∣≥0; ∣u∣=0 exactly when u=0; ∣u∣=u for u≥0; the order is total and trichotomy holds; and 0<1, so 0≠1 (Basic properties of the absolute value, The multiplicative identity is positive, Ordered field).

[L5]

Intervals: [1,2]={ y:1≤y≤2 } (Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Refutation

technique · direct
1.1

The point 0 lies in A, and N1(0)∩A={0}: an element of A is either 0, which satisfies ∣0∣=0<1, or an element of [1,2], which satisfies ∣y∣=y≥1 and so is not in N1(0). Hence 0 is an isolated point of A and not a limit point of A.

L2L3L4L5
1.2

The reals 0 and 1 are distinct.

L4
2.1

Take δ:=1. No x∈A satisfies 0<∣x−0∣<1: such an x would lie in N1∗(0)∩A, which is contained in N1(0)∩A={0} and excludes 0, hence is empty. So for every real L and every real ε>0 the choice δ=1 makes the implication in (∗) vacuously true, and every real L satisfies (∗) at c=0.

step 1.1L1L3L4
3.1

In particular L=0 and L=1 both satisfy (∗) at c=0, and they are distinct: more than one real satisfies the formula, so the claim is false.

step 1.2step 2.1L4∎

Remarks

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: f<g near c implies lim⁡f<lim⁡g

Statement

False claim: let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R), let f,g:A→R have limits at c (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A), and suppose there is a real η>0 with

f(x)<g(x)for every x∈A with 0<∣x−c∣<η.

Then lim⁡x→cf(x)<lim⁡x→cg(x).

What is true is the non-strict version, If f≤g on a punctured neighbourhood of c then lim⁡f≤lim⁡g, non-strictly: the hypothesis f≤g near c gives lim⁡f≤lim⁡g, and that conclusion cannot be improved even when the hypothesis is strengthened to a strict inequality at every point.

Why the strengthening fails. Strictness at each point is not a uniform statement: it says g(x)−f(x)>0 for every x near c, with no lower bound on that positive quantity. The limit only sees the limit of g−f, and a function that is positive everywhere may have limit 0. What does survive is the uniform version: if g(x)−f(x)≥κ near c for a fixed real κ>0, then lim⁡g−lim⁡f≥κ>0, by applying If f≤g on a punctured neighbourhood of c then lim⁡f≤lim⁡g, non-strictly to f+κ and g.

Facts & Assumptions

Given: The set A:=R, the point c:=0, the constant function f:R→R with f(x):=0 for every x, and the function g:R→R with g(x):=∣x∣.

[L1]

The limit condition (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A): lim⁡x→ch(x)=P means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−P∣<ε.

[L3]

Absolute value: ∣u∣≥0; ∣u∣=0 exactly when u=0; and ∣u∣=u for u≥0, so ∣∣x∣−0∣=∣x∣ (Basic properties of the absolute value).

[L4]

Order in R: trichotomy, so ∣u∣≥0 together with ∣u∣≠0 gives ∣u∣>0, and t<t is impossible (Ordered field).

Refutation

technique · direct
1.1

The point 0 is a limit point of R.

L2
1.2

The strict hypothesis holds with η:=1: every x∈R with 0<∣x−0∣<1 has ∣x∣≠0, hence ∣x∣>0=f(x), that is f(x)<g(x).

L3L4
2.1

Both limits exist and are equal to 0. For f: ∣f(x)−0∣=0<ε for every x and every real ε>0, any δ serving. For g: given a real ε>0 take δ:=ε; every x with 0<∣x−0∣<δ satisfies ∣∣x∣−0∣=∣x∣<ε.

step 1.1L1L3
3.1

So f(x)<g(x) throughout a punctured neighbourhood of 0 while lim⁡x→0f(x)=0=lim⁡x→0g(x); the asserted strict inequality 0<0 is impossible by trichotomy, so the claim is false.

step 1.2step 2.1L4∎

Remarks

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

FALSE: a function with a limit at c is bounded on its whole domain

Statement

False claim: let A⊆R, let c be a limit point of A (Limit point, isolated point, adherent point, derived set, and dense subset of R) and let f:A→R have a limit at c (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A). Then f is bounded on A, that is, the image f(A) is a bounded subset of R (Lower bound, bounded below, bounded set).

What is true is the local statement, If f has a finite limit at c then f is bounded on some punctured neighbourhood of c: there is a radius δ>0 such that f is bounded on A∩Nδ∗(c). The radius is produced by the limit condition at the single tolerance ε=1, and it carries no information whatever about the values of f far from c, which the limit condition never constrains.

The witness below is f(x)=1/x on (0,∞) at the point c=1: the limit there is 1, and f is bounded near 1, while on the whole domain f takes values above every real.

Facts & Assumptions

Given: The set A:=(0,∞) (Intervals of R: the nine order-convex forms, nondegeneracy, and length), the point c:=1, and the function f:A→R with f(x):=1/x=x−1.

[L1]

The limit condition (The ε-δ limit lim⁡x→cf(x)=L of f:A→R at a limit point c of A): lim⁡x→ch(x)=P means that for every real ε>0 there is a real δ>0 such that every x in the domain of h with 0<∣x−c∣<δ satisfies ∣h(x)−P∣<ε.

[L3]

Absolute value: ∣u∣≥0; ∣u∣=u for u≥0; ∣uv∣=∣u∣ ∣v∣; ∣−u∣=∣u∣; and for t>0, ∣u∣<t is equivalent to −t<u<t (Basic properties of the absolute value).

[L4]

Inverses and order: a>0 gives a−1>0, and 0<a<b gives 0<b−1<a−1 (Inverses of positives are positive, and reciprocation reverses order); (a−1)−1=a for a≠0, inverses being unique (Field); and for t>0, u<v is equivalent to ut<vt (Sign rules for products and monotonicity of multiplication).

[L5]

Order arithmetic: 0<1, hence 2>0 and 1/2>0 with 1−1/2=1/2 (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities); of two positive reals the smaller is positive, the order being total (Ordered field).

[L6]

Archimedean property: for every real M there is a natural n≥1 with M<n⋅1R, and the canonical naturals satisfy n⋅1R>0 (Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing, Complete ordered field (least-upper-bound property)).

[L7]

Bounded set: S⊆R is bounded when it has an upper bound and a lower bound; a set with no upper bound is not bounded (Lower bound, bounded below, bounded set).

Refutation

technique · direct
1.1

The point 1 lies in A=(0,∞) and is a limit point of A: given a real ε>0, let ρ be the smaller of ε and 1, so ρ>0; then 1+ρ/2>1>0 lies in A and satisfies 0<∣(1+ρ/2)−1∣=ρ/2<ε.

L2L3L5
1.2

f is well defined on A: every x∈A has x>0, hence x≠0 and x−1 exists, with x−1>0.

L4L7
2.1

The limit of f at 1 exists and equals 1. Let ε>0 be an arbitrary real and let δ be the smaller of 1/2 and ε/2, so δ>0. For x∈A with 0<∣x−1∣<δ we get x>1−1/2=1/2>0, hence 0<1/x<2 by [L4]; and ∣1/x−1∣=∣(1−x)/x∣=∣x−1∣⋅(1/x)<δ⋅2≤ε.

step 1.1step 1.2L1L3L4L5
2.2

The image f(A) has no upper bound. Let M be an arbitrary real; by [L6] fix a natural n≥1 with M<n⋅1R, and note n⋅1R>0. Then x:=(n⋅1R)−1 satisfies x>0, so x∈A, and f(x)=x−1=n⋅1R>M. So no real bounds f(A) above, and f(A) is not bounded.

step 1.2L4L6L7
3.1

So f has a limit at the limit point c=1 of its domain and is unbounded on that domain: the claim is false, while If f has a finite limit at c then f is bounded on some punctured neighbourhood of c remains true and gives boundedness on A∩N1/2∗(1), where indeed 0<f(x)<2 by step 2.1.

step 2.1step 2.2L7∎

Remarks

  • The limit hypothesis is entirely local and the conclusion asked for is global, so no argument could bridge them. The witness makes that concrete by putting the unbounded behaviour at the other end of the domain, arbitrarily far from c in the only sense available here.

  • The sequential analogue is true, and that contrast is worth noting: a convergent sequence is bounded (Every convergent sequence is bounded), because a sequence has only finitely many terms outside any tail, and finitely many reals are bounded. A function has no such structure: the part of A outside a punctured neighbourhood of c can be infinite and can carry arbitrary values.

  • A bounded version does hold with an extra hypothesis: if A is itself contained in a punctured neighbourhood of c on which the limit estimate applies, then local and global boundedness coincide. That is a hypothesis on the domain, not a theorem about limits.

Sources