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.

✓ 11 results · all verified · 9 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 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Limits of Real Functions: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

RemarkRemark: AI-generatedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-09 (gpt-5.6-terra-codex-subscription)Open item page →

The classical form of the oscillator above is $\sin (1 / x)$ , which this library can only construct much later

Orientation, not a claim of this library

Every analysis course states the two examples of this page in the form

$\sin (1 / x) has no limit at 0, x \sin (1 / x) \to 0 as x \to 0,$

and a reader who has met them before will recognise $ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$ and $x ψ (1 / x) \to 0$ as $x \to 0$ , by the squeeze theorem as those examples with $ψ$ in place of $\sin$ . This remark records the correspondence, and it records that the correspondence is orientation only: the two displayed statements are reported as what the classical treatment proves, not asserted here, and nothing on this page uses or proves anything about $\sin$ .

The later analytic construction

This library now constructs sine and cosine from their power series, proves their differential and addition laws, and defines pi from the first positive zero of cosine. Under this library's counterexample convention, sin(1/x) has no limit as x tends to zero ↗ displays the false proposition that the sine limit exists under Statement refuted, then proves it false; x sin(1/x) tends to zero despite its oscillation ↗ proves the squeezed limit for the product. Both occur later in the reading order, so the links are orientation-only forward references declared in this item's forward_refs; no proof on this earlier page depends on them.

What $\psi$ supplies instead

The function $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ of The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic is elementary — it needs only the integer part, the order and the absolute value — and it has the three properties that make the classical examples work:

it is bounded, with values exactly in $[0, 1 / 2]$ ;
it is periodic, with period $1$ , so $ψ (1 / x)$ oscillates without damping as $x \to 0$ ;
it attains two distinct values on every punctured neighbourhood of $0$ after the substitution $x \mapsto 1 / x$ , namely $0$ at the reciprocals of the integers and $1 / 2$ at the reciprocals of the half-integers.

The third property is what $ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$ uses. It is not sharper than what $\sin$ would give — the classical witnessing sequences hit the extreme values of $\sin$ exactly too — but it is available here: the two values $0$ and $1 / 2$ are read off from the integer part in one line (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ), with no series and no $π$ , whereas the corresponding facts about $\sin$ presuppose the whole construction described above.

What is genuinely lost, and what is not

Nothing on this page is weaker for using $ψ$ . The two statements proved are exactly the statements usually proved with $\sin$ , and their proofs are shorter.

What is lost is a connection to a different subject. The classical pair $\sin (1 / x)$ , $x \sin (1 / x)$ also carries information about smoothness, about power series and about the topologist's sine curve, none of which $ψ$ can carry, since $ψ$ is assembled from the order, the absolute value and the integer part alone. Those notions occur only later in the reading order and are unavailable on this earlier page.

5 · Examples, counterexamples and false statements

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

Every polynomial has $\lim_{x \to c} p (x) = p (c)$ , and rational functions do so away from the zeros of the denominator

Example

For a list $a_{0}, \dots, a_{n - 1}$ of reals write

$p (x) : = \sum_{j < n} a_{j} x^{j}$

for the finite sum of Finite sums and finite products, by recursion applied to the list $j \mapsto a_{j} x^{j}$ , with powers as in Integer powers $a^{m}$ . So $\sum_{j < 0}$ is the empty sum $0$ , and $p$ is a function $R \to R$ ; these are the polynomial functions.

Claim 1. For every polynomial function $p$ and every $c \in R$ , the limit of $p$ at $c$ exists and

$\lim_{x \to c} p (x) = p (c) .$

Claim 2. Let $p$ and $q$ be polynomial functions and let $c \in R$ satisfy $q (c) \neq 0$ . Put $A_{q} : = {x \in R : q (x) \neq 0}$ . Then $c \in A_{q}$ , the point $c$ is a limit point of $A_{q}$ , the quotient $p / q$ is defined on $A_{q}$ , its limit at $c$ exists, and

$\lim_{x \to c} (p / q) ∣_{A_{q}} (x) = \frac{p (c)}{q (c)} .$

Everything is read off from Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero once two trivial limits are in hand: that of a constant function and that of the identity. Note that claim 1 is exactly the statement that $\lim_{x \to c} p (x) = p (c)$ , the equality that FALSE: $\lim_{x \to c} f (x) = f (c)$ whenever both sides exist shows is not automatic; for polynomials it is a theorem, and the algebra of limits is what proves it.

Facts & Assumptions

Given: A list $a_{0}, \dots, a_{n - 1}$ of reals and the polynomial function $p (x) = \sum_{j < n} a_{j} x^{j}$ ; a second polynomial function $q$ ; and a real $c$ (Finite sums and finite products, by recursion, Integer powers $a^{m}$ ).

[L1]

The limit condition (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ): $\lim_{x \to c} h (x) = P$ 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 $∣ h (x) - P ∣ < ε$ .

[L2]

Algebra of function limits: at a limit point $c$ of the common domain, the limits of $f + g$ , of $α f$ and of $f g$ exist and equal $L + M$ , $α L$ and $L M$ ; and if $M \neq 0$ the limit of $(f / g)$ restricted to ${g \neq 0}$ exists and equals $L / M$ (Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero).

[L3]

Every real is a limit point of $R$ , punctured neighbourhoods being 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 $R$ ).

[L4]

Finite sums: $\sum_{j < 0} a_{j} = 0$ and $\sum_{j < n + 1} a_{j} = (\sum_{j < n} a_{j}) + a_{n}$ (Finite sums and finite products, by recursion).

[L5]

Powers: $x^{0} = 1$ and $x^{j + 1} = x^{j} \cdot x$ for every $x \in R$ and $j \in N$ (Integer powers $a^{m}$ ).

[L6]

Induction principle on $N$ (The principle of mathematical induction).

[L7]

Sign preservation: if the limit of $g$ at $c$ is nonzero then $c$ is a limit point of ${x : g (x) \neq 0}$ (If $\lim_{x \to c} f (x) = L \neq 0$ then $∣ f ∣ > ∣ L ∣ / 2$ on a punctured neighbourhood of $c$ ; in particular if $L > 0$ then $f > L / 2 > 0$ there).

[L8]

Absolute value: $∣ 0 ∣ = 0$ ; and field arithmetic (Basic properties of the absolute value, Field).

Verification

technique · direct

1.1

Every $c \in R$ is a limit point of $R$ , so [L1] and [L2] apply at $c$ to functions defined on $R$ .

1.2

A constant function $x \mapsto a$ has limit $a$ at $c$ : $∣ a - a ∣ = ∣ 0 ∣ = 0 < ε$ for every real $ε > 0$ , any $δ$ serving.

L1L8

1.3

The identity function $x \mapsto x$ has limit $c$ at $c$ : given a real $ε > 0$ , take $δ : = ε$ ; then $0 < ∣ x - c ∣ < δ$ gives $∣ x - c ∣ < ε$ .

2.1

For every $j \in N$ the function $x \mapsto x^{j}$ has limit $c^{j}$ at $c$ . This is an induction on $j$ [L6]. For $j = 0$ the function is the constant $1$ by [L5], and step 1.2 applies with $a = 1 = c^{0}$ . If the claim holds for $j$ , then $x^{j + 1} = x^{j} \cdot x$ by [L5], and the product rule of [L2] applied to $x \mapsto x^{j}$ and the identity gives limit $c^{j} \cdot c = c^{j + 1}$ .

step 1.2step 1.3L2L5L6

3.1

For every $j \in N$ the function $x \mapsto a_{j} x^{j}$ has limit $a_{j} c^{j}$ at $c$ , by the scalar rule of [L2] applied to step 2.1 with $α = a_{j}$ .

step 2.1L2

4.1

For every $n \in N$ the function $x \mapsto \sum_{j < n} a_{j} x^{j}$ has limit $\sum_{j < n} a_{j} c^{j}$ at $c$ . This is an induction on $n$ [L6]. For $n = 0$ both the function and the asserted limit are the empty sum $0$ by [L4], and step 1.2 applies. If the claim holds for $n$ , then $\sum_{j < n + 1} a_{j} x^{j} = (\sum_{j < n} a_{j} x^{j}) + a_{n} x^{n}$ by [L4], and the sum rule of [L2] applied to the inductive hypothesis and step 3.1 gives limit $(\sum_{j < n} a_{j} c^{j}) + a_{n} c^{n} = \sum_{j < n + 1} a_{j} c^{j}$ . Taking the given $n$ , the limit of $p$ at $c$ exists and equals $p (c)$ : claim 1.

step 1.2step 3.1L2L4L6

5.1

Now let $q$ be a polynomial function with $q (c) \neq 0$ and put $A_{q} : = {x \in R : q (x) \neq 0}$ . By step 4.1 the limit of $q$ at $c$ exists and equals $q (c) \neq 0$ , so [L7] gives that $c$ is a limit point of $A_{q}$ ; and $c \in A_{q}$ because $q (c) \neq 0$ .

step 4.1L7

6.1

The quotient rule of [L2], applied on $R$ to $p$ and $q$ with $\lim_{x \to c} q (x) = q (c) \neq 0$ , gives that the limit of $(p / q) ∣_{A_{q}}$ at $c$ exists and equals $p (c) / q (c)$ : claim 2.

step 1.1step 4.1step 5.1L2∎

Remarks

Two inductions, and why they are separate. The first builds the monomials from the identity by repeated multiplication; the second builds the polynomial from the monomials by repeated addition. Each is an induction on the recursion clause of the object it builds (Integer powers $a^{m}$ and Finite sums and finite products, by recursion respectively), and neither can be replaced by dots.
Index hygiene. The sum is written $\sum_{j < n}$ , whose first index is $j = 0$ and whose empty case $n = 0$ is the zero function; the base case of step 4.1 is that empty case, and $x^{0} = 1$ holds for every real $x$ including $0$ (Integer powers $a^{m}$ ), so no index or value is left undefined.
What claim 2 does not say. It says nothing at a zero of $q$ . There the quotient is undefined, and whether it has a limit depends on $p$ as well. It may have one: on ${x : x \neq 0}$ the quotient $x^{2} / x$ equals $x$ , whose limit at $0$ is $0$ by step 1.3 and 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. It may also fail to have one. Nothing on this page decides such cases in general.

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

The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic

Example

Identify $Z$ with its canonical copy in $R$ (The integers as equivalence classes of pairs of naturals, The integers embed in the rationals, The rationals embed densely in the reals) and for $x \in R$ put

$D (x) : = {∣ x - n ∣ : n \in Z}, ψ (x) : = \inf D (x)$

(Greatest lower bound (infimum)). Write $m : = ⌊ x ⌋$ for the integer part of $x$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ) and $t : = x - m$ , so $0 \leq t < 1$ . Then:

Existence and attainment. $ψ (x)$ exists and is attained: $ψ (x) = \min {t, 1 - t} = \min {∣ x - m ∣, ∣ x - (m + 1) ∣},$ so $ψ (x) = ∣ x - n ∣$ for $n = m$ or $n = m + 1$ , and $ψ (x) = \min D (x)$ (Maximum and minimum of a set).
Range. $0 \leq ψ (x) \leq 1 / 2$ for every real $x$ , and every value in $[0, 1 / 2]$ occurs: the range of $ψ$ is exactly the interval $[0, 1 / 2]$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).
Zero set. $ψ (x) = 0$ if and only if $x \in Z$ .
Half-integers. $ψ (m + 1 / 2) = 1 / 2$ for every $m \in Z$ .
Periodicity. $ψ (x + 1) = ψ (x)$ for every real $x$ .

What this function is for. It is the elementary, trigonometry-free substitute for $\sin$ : it is bounded, it oscillates, and on every punctured neighbourhood of $0$ the composite $ψ (1 / x)$ attains both the value $0$ and the value $1 / 2$ . Claims 3 and 4 are exactly what the companion counterexample $ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$ evaluates, and claim 2 is what the squeeze argument of $x ψ (1 / x) \to 0$ as $x \to 0$ , by the squeeze theorem uses.

Facts & Assumptions

Given: A real $x$ ; the set $D (x) = {∣ x - n ∣ : n \in Z}$ ; the integer $m : = ⌊ x ⌋$ and the real $t : = x - m$ . Integers are identified with their canonical copies in $R$ .

[L1]

Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ). Hence $0 \leq t < 1$ and $0 < 1 - t \leq 1$ , where $1 - t = (m + 1) - x$ .

[L2]

Integers in $R$ : the embeddings $N \to Z \to Q \to R$ are injective and preserve $0$ , $1$ , addition and order; $Z$ is a totally ordered commutative ring, closed under $n \mapsto n + 1$ and $n \mapsto n - 1$ ; every integer $\geq 0$ is the image of a unique natural; and a natural $j \neq 0$ satisfies $j \geq 1$ , so an integer $> 0$ is $\geq 1$ and consequently, for integers $n < n^{'}$ , one has $n + 1 \leq n^{'}$ (The integers as equivalence classes of pairs of naturals, The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, The integers form a totally ordered ring, The integers form a commutative ring, Discreteness: $σ (n)$ is the immediate successor, The natural numbers $N$ (von Neumann)).

[L3]

Infimum: $ℓ = \inf S$ when $ℓ \leq s$ for every $s \in S$ and $ℓ^{'} \leq ℓ$ for every lower bound $ℓ^{'}$ of $S$ . So a lower bound of $S$ that belongs to $S$ is the infimum, and is then also the minimum of $S$ (Greatest lower bound (infimum), Maximum and minimum of a set, Lower bound, bounded below, bounded set).

[L4]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ exactly when $u = 0$ ; $∣ u ∣ = u$ for $u \geq 0$ and $∣ u ∣ = - u$ for $u \leq 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L5]

Order and field arithmetic in $R$ : the order is total and trichotomy holds; translation invariance and adding inequalities (Order is preserved by adding a constant and by adding inequalities); $0 < 1$ (The multiplicative identity is positive), so $2 > 0$ , $1 / 2 > 0$ (Inverses of positives are positive, and reciprocation reverses order), $1 / 2 < 1$ and $1 - 1 / 2 = 1 / 2$ (Sign rules for products and monotonicity of multiplication, Field); and the minimum of a two-element set of reals (Maximum and minimum of a set, Ordered field).

Verification

technique · direct

1.1

$D (x)$ is nonempty and $0$ is a lower bound of it: the integer $0$ gives $∣ x - 0 ∣ \in D (x)$ , and $∣ x - n ∣ \geq 0$ for every $n \in Z$ .

L2L4

1.2

By [L1] the integer $m = ⌊ x ⌋$ satisfies $m \leq x < m + 1$ , so $t = x - m$ satisfies $0 \leq t < 1$ , and $(m + 1) - x = 1 - t$ satisfies $0 < 1 - t \leq 1$ .

L1L5

2.1

Every element of $D (x)$ is at least $\min {t, 1 - t}$ . Let $n \in Z$ . By [L2] and totality either $n \leq m$ or $m < n$ , and in the second case $m + 1 \leq n$ . If $n \leq m$ then $x - n \geq x - m = t \geq 0$ , so $∣ x - n ∣ = x - n \geq t$ . If $m + 1 \leq n$ then $n - x \geq (m + 1) - x = 1 - t > 0$ , so $∣ x - n ∣ = n - x \geq 1 - t$ . In both cases $∣ x - n ∣ \geq \min {t, 1 - t}$ .

step 1.2L2L4L5

2.2

Both $t$ and $1 - t$ belong to $D (x)$ : since $t \geq 0$ we have $t = ∣ x - m ∣$ , and since $1 - t > 0$ we have $1 - t = ∣ x - (m + 1) ∣$ , with $m$ and $m + 1$ in $Z$ .

step 1.2L2L4

3.1

Hence $\min {t, 1 - t}$ is a lower bound of $D (x)$ belonging to $D (x)$ , so by [L3] it is the greatest lower bound and also the minimum: $ψ (x) = \min {t, 1 - t} = \min {∣ x - m ∣, ∣ x - (m + 1) ∣}$ , attained at $n = m$ or at $n = m + 1$ . This is claim 1.

step 2.1step 2.2L3L5

4.1

Claim 2, the inclusion. $ψ (x) \geq 0$ , since $t \geq 0$ and $1 - t > 0$ ; and $ψ (x) \leq 1 / 2$ : if $t \leq 1 / 2$ then $ψ (x) \leq t \leq 1 / 2$ , while if $1 / 2 < t$ then $1 - t < 1 - 1 / 2 = 1 / 2$ and $ψ (x) \leq 1 - t < 1 / 2$ . So $0 \leq ψ (x) \leq 1 / 2$ for every real $x$ .

step 1.2step 3.1L5

4.2

Claim 3. If $ψ (x) = 0$ then $\min {t, 1 - t} = 0$ ; since $1 - t > 0$ this forces $t = 0$ , that is $x = m \in Z$ . Conversely if $x \in Z$ then $∣ x - x ∣ = 0$ lies in $D (x)$ and $0$ is a lower bound of $D (x)$ by step 1.1, so $ψ (x) = 0$ by [L3].

step 1.1step 1.2step 3.1L3L4L5

4.3

Claim 4. Let $m \in Z$ and $x : = m + 1 / 2$ . Since $0 < 1 / 2 < 1$ we have $m \leq x < m + 1$ , so the uniqueness in [L1] gives $⌊ x ⌋ = m$ and $t = 1 / 2$ ; then step 3.1 gives $ψ (x) = \min {1 / 2, 1 - 1 / 2} = \min {1 / 2, 1 / 2} = 1 / 2$ .

step 3.1L1L5

4.4

Claim 5. The map $n \mapsto n + 1$ is a bijection of $Z$ onto itself, with inverse $n \mapsto n - 1$ [L2]; so, substituting $n = n^{'} + 1$ , $D (x + 1) = {∣ (x + 1) - n ∣ : n \in Z} = {∣ x - n^{'} ∣ : n^{'} \in Z} = D (x) .$ Being infima of the same set, $ψ (x + 1)$ and $ψ (x)$ are equal by step 3.1 applied at $x + 1$ and at $x$ .

step 3.1L2L3

5.1

Claim 2, the exact range. Every value of $ψ$ lies in $[0, 1 / 2]$ by step 4.1. Conversely let $s$ satisfy $0 \leq s \leq 1 / 2$ ; then $0 \leq s < 1$ , so $0 \leq s < 0 + 1$ and the uniqueness in [L1] gives $⌊ s ⌋ = 0$ and $t = s$ ; and $s \leq 1 / 2 \leq 1 - s$ because $2 s \leq 1$ , so step 3.1 gives $ψ (s) = \min {s, 1 - s} = s$ . Hence the range of $ψ$ is exactly $[0, 1 / 2]$ .

step 3.1step 4.1L1L5

6.1

So $ψ$ is defined at every real, is attained at a nearest integer, has range exactly $[0, 1 / 2]$ , vanishes exactly on $Z$ , takes the value $1 / 2$ at every half-integer, and is $1$ -periodic.

step 3.1step 4.1step 4.2step 4.3step 4.4step 5.1∎

Remarks

No completeness of $R$ is needed for the infimum here. The general existence theorem Every nonempty set bounded below has an infimum would supply $\inf D (x)$ from the least-upper-bound property, but step 3.1 does not use it: the infimum is produced by exhibiting an element of $D (x)$ that is also a lower bound, which is Greatest lower bound (infimum) read directly. Completeness does enter, once, through Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ , whose existence half is the Archimedean property.
Why $\min {t, 1 - t}$ and not "the distance to the nearest integer". The phrase presupposes that a nearest integer exists, which is exactly what step 2.2 establishes and what the picture cannot. When $t = 1 / 2$ there are two nearest integers, $m$ and $m + 1$ , and the formula is indifferent to which is chosen, so nothing has to be selected.
$ψ$ is the triangle wave of amplitude $1 / 2$ and period $1$ — not the sawtooth $x - ⌊ x ⌋$ , which drops discontinuously at every integer: on $[0, 1 / 2]$ it is $ψ (s) = s$ by step 5.1, and periodicity and the reflection $ψ (- x) = ψ (x)$ — immediate from $D (- x) = {∣ - x - n ∣ : n \in Z} = {∣ x + n ∣ : n \in Z} = D (x)$ , using $∣ - u ∣ = ∣ u ∣$ and the bijection $n \mapsto - n$ of $Z$ — determine it everywhere.

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

$x ψ (1 / x) \to 0$ as $x \to 0$ , by the squeeze theorem

Example

Let $A : = R ∖ {0}$ and define $h : A \to R$ by

$h (x) : = x \cdot ψ (1 / x),$

with $ψ$ as in The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic. Then $0$ is a limit point of $A$ , the limit of $h$ at $0$ exists, and

$\lim_{x \to 0} h (x) = 0.$

The point of the example. The factor $ψ (1 / x)$ has no limit at $0$ at all ( $ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$ ), so Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero cannot be applied to the product: its product rule requires both factors to have limits. What is available is that $ψ (1 / x)$ stays inside $[0, 1 / 2]$ , and a bounded factor multiplied by one tending to $0$ is killed. That is exactly what If $f \leq g \leq h$ near $c$ and $f$ and $h$ have the same limit at $c$ , then so does $g$ delivers, and it delivers the existence of the limit, not merely its value.

Facts & Assumptions

Given: The set $A = R ∖ {0}$ and the function $h : A \to R$ , $h (x) = x \cdot ψ (1 / x)$ , with $ψ$ the function of The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic.

[L1]

Range of $ψ$ : $0 \leq ψ (u) \leq 1 / 2$ for every real $u$ (The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic, claim 2).

[L2]

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

[L3]

Squeeze theorem: if $f \leq g \leq k$ on $A \cap N_{η}^{*} (c)$ for some real $η > 0$ , and the limits of $f$ and of $k$ at $c$ exist and are equal to $L$ , then the limit of $g$ at $c$ exists and equals $L$ (If $f \leq g \leq h$ near $c$ and $f$ and $h$ have the same limit at $c$ , then so does $g$ ).

[L4]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ exactly when $u = 0$ ; $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ ; $∣ - u ∣ = ∣ u ∣$ ; $∣ u ∣ = u$ for $u \geq 0$ ; and $- ∣ u ∣ \leq u \leq ∣ u ∣$ (Basic properties of the absolute value).

[L5]

Order and field arithmetic: $x \neq 0$ has an inverse $1 / x$ (Field); $0 < 1$ , so $2 > 0$ and $1 / 2 > 0$ with $t / 2 < t$ for $t > 0$ (The multiplicative identity is positive, Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication); multiplying an inequality by a non-negative factor, and adding inequalities (Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities); the order is total (Ordered field). Those two sources state their moves in their STRICT forms only; the non-strict forms used below follow by adjoining the equality case, in which the two sides coincide (Ordered field).

[L6]

Limit point and neighbourhoods (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 $R$ ).

Verification

technique · direct

1.1

$0$ is a limit point of $A = R ∖ {0}$ : given a real $ε > 0$ , the real $ε / 2$ satisfies $ε / 2 > 0$ , so it lies in $A$ , and $0 < ∣ ε / 2 - 0 ∣ = ε / 2 < ε$ .

L4L5L6

1.2

$h$ is defined on all of $A$ , and $∣ h (x) ∣ \leq ∣ x ∣ / 2$ there: for $x \in A$ we have $x \neq 0$ , so $1 / x$ exists, and $∣ h (x) ∣ = ∣ x ∣ \cdot ψ (1 / x)$ by [L4], while $0 \leq ψ (1 / x) \leq 1 / 2$ by [L1] and $∣ x ∣ \geq 0$ , so multiplying the inequality $ψ (1 / x) \leq 1 / 2$ by the non-negative factor $∣ x ∣$ gives $∣ h (x) ∣ \leq ∣ x ∣ / 2$ .

L1L4L5

1.3

The two functions $x \mapsto - ∣ x ∣ / 2$ and $x \mapsto ∣ x ∣ / 2$ on $A$ each have limit $0$ at $0$ : given a real $ε > 0$ , take $δ : = ε$ ; every $x \in A$ with $0 < ∣ x - 0 ∣ < δ$ satisfies $∣ ∣ x ∣ / 2 - 0 ∣ = ∣ x ∣ / 2 < ε / 2 < ε$ , and likewise $∣ - ∣ x ∣ / 2 - 0 ∣ = ∣ x ∣ / 2 < ε$ .

L2L4L5

2.1

Hence $- ∣ x ∣ / 2 \leq h (x) \leq ∣ x ∣ / 2$ for every $x \in A$ , by [L4] applied to $∣ h (x) ∣ \leq ∣ x ∣ / 2$ .

step 1.2L4

3.1

The three functions satisfy $- ∣ x ∣ / 2 \leq h (x) \leq ∣ x ∣ / 2$ on all of $A$ , in particular on $A \cap N_{1}^{*} (0)$ , and the two outer ones have limit $0$ at $0$ ; since $0$ is a limit point of $A$ , the squeeze theorem [L3] gives that the limit of $h$ at $0$ exists and equals $0$ .

step 1.1step 1.3step 2.1L3∎

Remarks

Where the hypotheses of the squeeze theorem are met. The order hypothesis holds on all of $A$ , so any $η > 0$ serves and $η = 1$ is taken; the two outer limits are computed by hand in step 1.3; and $0$ is a limit point of $A$ by step 1.1, which is what makes every limit here well posed (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).
Nothing about $ψ$ beyond its range is used. Replacing $ψ$ by any function with values in a fixed bounded set would give the same conclusion by the same three steps. What makes the example worth stating is the contrast with $ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$ : the same oscillating factor, multiplied by $x$ or not, is the difference between a limit existing and not.
The classical version of this example is $x \sin (1 / x) \to 0$ as $x \to 0$ ; see The classical form of the oscillator above is $\sin (1 / x)$ , which this library can only construct much later for why this library writes $ψ$ and not $\sin$ .

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

The sign function has both one-sided limits at $0$ and no two-sided limit

Example

Define $sgn : R \to R$ by

$sgn (x) : = {\begin{cases} 1 & x > 0, \\ 0 & x = 0, \\ - 1 & x < 0. \end{cases}$

Then $0$ is a limit point of both $(0, \infty)$ and $(- \infty, 0)$ , both one-sided limits at $0$ exist (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ ),

$\lim_{x \to 0^{+}} sgn (x) = 1, \lim_{x \to 0^{-}} sgn (x) = - 1,$

and $sgn$ has no limit at $0$ .

This is the standard illustration of If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree: the two one-sided limits both exist, so nothing is missing on either side, yet they disagree, and disagreement is exactly what the theorem converts into the failure of the two-sided limit. Note also that the value $sgn (0) = 0$ is equal to neither one-sided limit, and is irrelevant to all three assertions (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

Facts & Assumptions

Given: The function $sgn : R \to R$ above, with $A : = R$ , $c : = 0$ , $A^{+} = R \cap (0, \infty) = (0, \infty)$ and $A^{-} = R \cap (- \infty, 0) = (- \infty, 0)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L1]

[L2]

One-sided limits are the limits at $c$ of the restrictions of $f$ to $A^{-}$ and $A^{+}$ , and are well posed exactly when $c$ is a limit point of the set in question (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ ).

[L3]

Limit point and neighbourhoods (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 $R$ ).

[L4]

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

[L5]

Order in $R$ : trichotomy, so every real satisfies exactly one of $x > 0$ , $x = 0$ , $x < 0$ ; $0 < 1$ and hence $2 > 0$ , $ε / 2 > 0$ and $ε / 2 < ε$ for $ε > 0$ ; and $1 - (- 1) = 1 + 1 > 0$ , so $- 1 < 1$ and in particular $1 \neq - 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Ordered field).

[L6]

Two-sided versus one-sided: if $c$ is a limit point of both $A^{-}$ and $A^{+}$ and $\lim_{x \to c} f (x) = L$ , then $\lim_{x \to c^{-}} f (x) = L$ and $\lim_{x \to c^{+}} f (x) = L$ (If $c$ is a limit point of the domain from both sides, the limit exists iff both one-sided limits exist and agree).

[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 the restrictions, each one-sided limit is a single real.

Verification

technique · direct

1.1

$sgn$ is a well-defined function on $R$ : by trichotomy every real satisfies exactly one of the three defining conditions.

1.2

$0$ is a limit point of $A^{+} = (0, \infty)$ and of $A^{-} = (- \infty, 0)$ : given a real $ε > 0$ , the real $ε / 2$ is positive, hence lies in $A^{+}$ , and satisfies $0 < ∣ ε / 2 - 0 ∣ = ε / 2 < ε$ ; and $- ε / 2$ is negative, hence lies in $A^{-}$ , and satisfies $0 < ∣ - ε / 2 - 0 ∣ = ε / 2 < ε$ .

L3L4L5

1.3

The reals $1$ and $- 1$ are distinct, since $1 - (- 1) = 1 + 1 > 0$ .

2.1

$\lim_{x \to 0^{+}} sgn (x) = 1$ : by [L2] this is the limit at $0$ of the restriction of $sgn$ to $A^{+}$ , which is well posed by step 1.2. Given a real $ε > 0$ , any $δ > 0$ serves, since every $x \in A^{+}$ has $x > 0$ , hence $sgn (x) = 1$ and $∣ sgn (x) - 1 ∣ = ∣ 0 ∣ = 0 < ε$ .

step 1.1step 1.2L1L2L4

2.2

$\lim_{x \to 0^{-}} sgn (x) = - 1$ : identically, every $x \in A^{-}$ has $x < 0$ , hence $sgn (x) = - 1$ and $∣ sgn (x) - (- 1) ∣ = ∣ 0 ∣ = 0 < ε$ for every $ε > 0$ and every $δ > 0$ .

step 1.1step 1.2L1L2L4

3.1

Suppose $sgn$ had a limit at $0$ , say $\lim_{x \to 0} sgn (x) = L$ . Since $0$ is a limit point of both $A^{-}$ and $A^{+}$ by step 1.2, [L6] gives $\lim_{x \to 0^{-}} sgn (x) = L$ and $\lim_{x \to 0^{+}} sgn (x) = L$ ; each one-sided limit is single valued by [L7], so steps 2.1 and 2.2 force $L = 1$ and $L = - 1$ , contradicting step 1.3. Hence $sgn$ has no limit at $0$ .

step 1.3step 2.1step 2.2L6L7∎

Remarks

The failure is not about the value at $0$ . Redefining $sgn (0)$ to be $1$ , or $- 1$ , or anything else changes nothing: The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ never evaluates the function at the point, and the two one-sided limits are computed on sets that exclude $0$ (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ ). This is a genuine jump, not a removable defect of the kind FALSE: $\lim_{x \to c} f (x) = f (c)$ whenever both sides exist exhibits.
Away from $0$ the function is locally constant, so it has a limit at every other point of $R$ , equal to its value there: for $c > 0$ take $δ$ to be $c$ itself, and every $x$ with $0 < ∣ x - c ∣ < c$ has $x > 0$ and $sgn (x) = 1 = sgn (c)$ ; symmetrically for $c < 0$ . So the single point $0$ carries the whole phenomenon.
Contrast with the two other failures on this page. Here both one-sided limits exist and differ; for $ψ (1 / x)$ at $0$ ( $ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$ ) the failure is already one-sided, both witnessing sequences there having positive terms; and for the indicator of $Q$ (The indicator of $Q$ has a limit at no point of $R$ ) the failure occurs at every point at once.

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

$(3 x^{2} - 1) / (x^{2} + x) \to 3$ as $x \to + \infty$

Example

Let $A : = (0, \infty)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) and let

$f : A \to R, f (x) : = \frac{3 x^{2} - 1}{x^{2} + x}$

(Integer powers $a^{m}$ ). Then $A$ is not bounded above (Lower bound, bounded below, bounded set), so the limit at $+ \infty$ is well posed (Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point); it exists, and

$\lim_{x \to + \infty} f (x) = 3.$

This is proved by a direct estimate, not by an algebra of limits. Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero is stated at a finite limit point of the domain, and this library proves no algebra of limits at $\pm \infty$ ; the familiar manipulation "divide numerator and denominator by $x^{2}$ and take limits termwise" is therefore not available here. Instead the whole computation is packed into one inequality, valid for $x \geq 1$ :

$∣ f (x) - 3 ∣ = \frac{1 + 3 x}{x^{2} + x} \leq \frac{4}{x},$

after which the Archimedean property finishes the argument.

Facts & Assumptions

Given: The set $A = (0, \infty)$ and the function $f (x) = (3 x^{2} - 1) / (x^{2} + x)$ on $A$ .

[L1]

Limits at $+ \infty$ : for $A$ not bounded above, $\lim_{x \to + \infty} f (x) = L$ means that for every real $ε > 0$ there is a real $M$ with $∣ f (x) - L ∣ < ε$ for every $x \in A$ with $x > M$ (Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point).

[L2]

Archimedean property: for every real $t$ there is a natural $n \geq 1$ with $t < n \cdot 1_{R}$ ; and for every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Complete ordered field (least-upper-bound property)). The canonical naturals satisfy $n \cdot 1_{R} > 0$ and $1 \leq n \cdot 1_{R}$ for $n \geq 1$ , and are increasing in $n$ (Canonical naturals are positive and strictly increasing).

[L3]

Bounded set: $S$ is bounded above when some real is an upper bound of it (Lower bound, bounded below, bounded set); and $(0, \infty) = {x : x > 0}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Order and field arithmetic: products of positives are positive and for $t > 0$ , $u < v$ is equivalent to $u t < v t$ (Sign rules for products and monotonicity of multiplication); $a > 0$ gives $a^{- 1} > 0$ and $0 < a < b$ gives $0 < 1 / b < 1 / a$ , with the non-strict forms following by adjoining equality (Inverses of positives are positive, and reciprocation reverses order); adding inequalities and translation invariance (Order is preserved by adding a constant and by adding inequalities); $0 < 1$ (The multiplicative identity is positive); the field identities (Field); transitivity and totality (Ordered field).

[L5]

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

[L6]

Powers: $x^{2} = x \cdot x$ (Integer powers $a^{m}$ ).

Verification

technique · direct

1.1

$f$ is defined on all of $A$ : every $x \in A$ has $x > 0$ , hence $x^{2} = x \cdot x > 0$ and $x^{2} + x > 0$ , so $x^{2} + x \neq 0$ and the quotient exists.

L3L4L6

1.2

$A$ is not bounded above: given a real $M$ , [L2] supplies a natural $n \geq 1$ with $M < n \cdot 1_{R}$ , and $n \cdot 1_{R} > 0$ puts it in $A$ ; so no real is an upper bound of $A$ , and the limit at $+ \infty$ is well posed.

L2L3

2.1

For every $x \in A$ , $f (x) - 3 = \frac{(3 x^{2} - 1) - 3 (x^{2} + x)}{x^{2} + x} = \frac{- 1 - 3 x}{x^{2} + x}$ , hence, both $1 + 3 x$ and $x^{2} + x$ being positive, $∣ f (x) - 3 ∣ = \frac{1 + 3 x}{x^{2} + x}$ .

step 1.1L4L5L6

3.1

For every $x \in A$ with $x \geq 1$ : from $1 \leq x$ we get $1 + 3 x \leq x + 3 x = 4 x$ , and from $x > 0$ we get $x^{2} + x > x^{2} > 0$ ; therefore $\frac{1 + 3 x}{x^{2} + x} \leq \frac{4 x}{x^{2} + x} \leq \frac{4 x}{x^{2}} = \frac{4}{x}$ , so $∣ f (x) - 3 ∣ \leq 4 / x$ .

step 2.1L4L6

4.1

Let $ε > 0$ be an arbitrary real. By [L2] fix a natural $n \geq 1$ with $1 / n < ε$ , and put $M : = 4 n$ , where $n$ denotes the canonical natural $n \cdot 1_{R}$ . Since $n \geq 1$ we have $M = 4 n \geq 4 > 1$ . For every $x \in A$ with $x > M$ : first $x > 1$ , so step 3.1 applies and $∣ f (x) - 3 ∣ \leq 4 / x$ ; and $0 < M < x$ gives $0 < 1 / x < 1 / M$ by [L4], whence $4 / x < 4 / M = 4 / (4 n) = 1 / n < ε$ . So $∣ f (x) - 3 ∣ < ε$ for every $x \in A$ with $x > M$ .

step 3.1L2L4L5

5.1

Since $A$ is not bounded above and for every real $ε > 0$ such an $M$ has been produced, the limit of $f$ at $+ \infty$ exists and equals $3$ .

step 1.2step 4.1L1∎

Remarks

Where the estimate comes from. The exact identity of step 2.1 replaces the informal "the leading terms dominate": it makes $∣ f (x) - 3 ∣$ a quotient of two explicit positive quantities, and step 3.1 then bounds numerator above and denominator below by the crudest possible expressions, $4 x$ and $x^{2}$ . The constant $4$ is not optimal and does not need to be: the Archimedean property absorbs any constant.
Why the domain is $(0, \infty)$ and not $R$ . The denominator $x^{2} + x$ vanishes at $0$ and at $- 1$ , so $f$ is not defined there; restricting to $(0, \infty)$ both makes $f$ a function and makes the denominator positive, which is what lets the absolute values be dropped in step 2.1. Any domain unbounded above and avoiding the two zeros would give the same limit by the same estimate.
The corresponding statement at $- \infty$ would be the limit $3$ on a domain unbounded below and avoiding the two zeros of the denominator, proved from the same identity of step 2.1 with the inequalities on $x$ reversed. It is not asserted here and is not proved here, because nothing on these pages uses it.

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

$ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$

Statement refuted

Refuted claim: the function

$Φ : R ∖ {0} \to R, Φ (x) : = ψ (1 / x),$

with $ψ$ the distance to the integers (The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic), has a limit at $0$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

$Φ$ is bounded — $0 \leq Φ (x) \leq 1 / 2$ for every $x \neq 0$ , by claim 2 of The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic — and $0$ is a limit point of its domain, so every hypothesis that might plausibly deliver a limit except the limit itself is present. Boundedness near a point is therefore not sufficient for a limit to exist, and the converse of If $f$ has a finite limit at $c$ then $f$ is bounded on some punctured neighbourhood of $c$ fails.

The refutation exhibits two sequences of positive reals tending to $0$ along which $Φ$ is constantly $0$ and constantly $1 / 2$ , and applies A function has no limit at $c$ as soon as two sequences in $A ∖ {c}$ tending to $c$ give different limits of the values.

Facts & Assumptions

Given: The function $Φ (x) = ψ (1 / x)$ on $R ∖ {0}$ , and the sequences $x_{k} : = 1 / (k + 1)$ and $y_{k} : = 2 / (2 k + 1)$ for $k \in N$ . 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 first terms are $x_{0} = 1$ and $y_{0} = 2$ ; the denominators $k + 1$ and $2 k + 1$ are canonical naturals $\geq 1$ , never $0$ , which is why the sequences are written this way and not as $1 / k$ .

[L1]

The function $ψ$ vanishes exactly on $Z$ , satisfies $ψ (m + 1 / 2) = 1 / 2$ for every $m \in Z$ , and takes values in $[0, 1 / 2]$ (The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic, claims 2, 3 and 4).

[L2]

Nonexistence criterion: if two sequences with all terms in $A ∖ {c}$ converge to $c$ while the image sequences converge to distinct reals, then $f$ has no limit at $c$ (A function has no limit at $c$ as soon as two sequences in $A ∖ {c}$ tending to $c$ give different limits of the values).

[L3]

Sequential convergence, and the fact that a constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences). Testing against every positive real rather than every positive rational defines the same relation (The rationals embed densely in the reals, remarks of Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L4]

Reciprocal Archimedean property: for every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean); canonical naturals are positive and strictly increasing in the index (Canonical naturals are positive and strictly increasing); and $0 < a < b$ gives $0 < 1 / b < 1 / a$ , with the non-strict form following by adjoining equality (Inverses of positives are positive, and reciprocation reverses order).

[L5]

Limit point and neighbourhoods (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 $R$ ).

[L6]

Absolute value (Basic properties of the absolute value); order and field arithmetic: $0 < 1$ , so $2 > 0$ and $1 / 2 > 0$ with $1 / 2 \neq 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Field, Ordered field).

[L7]

Integers in $R$ : every canonical natural is an integer, and $Z$ is closed under adding $1$ (The integers as equivalence classes of pairs of naturals, The naturals embed in the integers, The integers embed in the rationals, The rationals embed densely in the reals, The integers form a commutative ring).

Counterexample

technique · direct

1.1

$0$ is a limit point of $R ∖ {0}$ : given a real $ε > 0$ , the real $ε / 2$ is positive, hence lies in $R ∖ {0}$ , and $0 < ∣ ε / 2 - 0 ∣ = ε / 2 < ε$ .

L5L6

1.2

For every $k \in N$ the terms $x_{k} = 1 / (k + 1)$ and $y_{k} = 2 / (2 k + 1)$ are defined and positive, since $k + 1 \geq 1 > 0$ and $2 k + 1 \geq 1 > 0$ ; in particular $x_{k} \neq 0$ and $y_{k} \neq 0$ , so both sequences have all their terms in $R ∖ {0}$ , which equals $(R ∖ {0}) ∖ {0}$ .

L4L6L7

1.3

The reals $0$ and $1 / 2$ are distinct, since $1 / 2 > 0$ .

2.1

$x_{k} \to 0$ : given a real $ε > 0$ , [L4] supplies a natural $n \geq 1$ with $1 / n < ε$ ; every $k \geq n$ has $k + 1 > n \geq 1$ , hence $∣ x_{k} - 0 ∣ = 1 / (k + 1) < 1 / n < ε$ .

step 1.2L3L4L6

2.2

$y_{k} \to 0$ : for every $k \in N$ we have $2 k + 1 \geq k + 1$ , since their difference is $k \geq 0$ , so $0 < y_{k} = 2 / (2 k + 1) \leq 2 / (k + 1)$ . Given a real $ε > 0$ , [L4] supplies a natural $n \geq 1$ with $1 / n < ε / 2$ ; every $k \geq n$ has $k + 1 > n$ , hence $∣ y_{k} - 0 ∣ \leq 2 / (k + 1) < 2 / n < ε$ .

step 1.2L3L4L6

2.3

$Φ (x_{k}) = 0$ for every $k$ : $1 / x_{k} = k + 1$ , a canonical natural and hence an integer by [L7], so $ψ (1 / x_{k}) = 0$ by [L1]. The image sequence is therefore the constant sequence $0$ and converges to $0$ .

step 1.2L1L3L7

2.4

$Φ (y_{k}) = 1 / 2$ for every $k$ : $1 / y_{k} = (2 k + 1) / 2 = k + 1 / 2$ with $k$ an integer by [L7], so $ψ (1 / y_{k}) = 1 / 2$ by [L1]. The image sequence is therefore the constant sequence $1 / 2$ and converges to $1 / 2$ .

step 1.2L1L3L6L7

3.1

So $(x_{k})$ and $(y_{k})$ have all their terms in $(R ∖ {0}) ∖ {0}$ and both converge to $0$ , which is a limit point of $R ∖ {0}$ , while the image sequences converge to the distinct reals $0$ and $1 / 2$ . By [L2], $Φ$ has no limit at $0$ .

step 1.1step 1.3step 2.1step 2.2step 2.3step 2.4L2∎

Remarks

Both witnessing sequences have positive terms, so what is refuted is already the existence of the right-hand limit $\lim_{x \to 0^{+}} Φ (x)$ (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ ), and the two-sided failure follows. The contrast with the sign function on this page is exact: there both one-sided limits exist and merely disagree.
Why $1 / (k + 1)$ and $2 / (2 k + 1)$ . They are the sequences whose reciprocals are $k + 1$ and $k + 1 / 2$ , that is, the integers and the half-integers, which are precisely the two sets on which claims 3 and 4 of The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic evaluate $ψ$ exactly. Writing $1 / k$ instead would be undefined at the index $k = 0$ , since $N$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).
Multiplying by $x$ repairs it. The function $x \mapsto x ψ (1 / x)$ does have a limit at $0$ , namely $0$ , by the squeeze theorem ( $x ψ (1 / x) \to 0$ as $x \to 0$ , by the squeeze theorem). The oscillation is unchanged; what changes is that its amplitude is forced to $0$ .
The classical form of this counterexample uses $\sin (1 / x)$ in place of $ψ (1 / x)$ ; The classical form of the oscillator above is $\sin (1 / x)$ , which this library can only construct much later records why this library cannot yet write it.

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

The function equal to $0$ off the origin and to $1$ at the origin has limit $0 \neq 1$ there

Statement refuted

Refuted claim: if $c \in A$ is a limit point of $A \subseteq R$ and $f : A \to R$ has a limit at $c$ , then $\lim_{x \to c} f (x) = f (c)$ — the false statement FALSE: $\lim_{x \to c} f (x) = f (c)$ whenever both sides exist.

The witness is the smallest one available: the function

$f : R \to R, f (x) : = {\begin{cases} 0 & x \neq 0, \\ 1 & x = 0, \end{cases}$

at the point $c = 0$ . It has limit $0$ there, while $f (0) = 1$ .

Beyond refuting the claim, this item records two further facts about the same witness, both used elsewhere on the page: both one-sided limits at $0$ also equal $0$ , so the defect is not a jump; and changing the single value $f (0)$ to $0$ produces a function with the same limit and the equality restored. That is what makes this a removable defect, and it is the pattern the composition counterexample With $g \equiv 0$ and $f$ equal to $0$ off the origin and $1$ at it, $\lim g = 0$ and $\lim_{y \to 0} f = 0$ while $f \circ g \equiv 1$ exploits.

Facts & Assumptions

Given: The function $f : R \to R$ above and the point $c : = 0$ ; and the constant function $g : R \to R$ with $g (x) : = 0$ for every $x$ .

[L1]

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

[L2]

Limit point: every real is a limit point of $R$ , punctured neighbourhoods being never empty; and $0$ is a limit point of $(0, \infty)$ and of $(- \infty, 0)$ , since $ε / 2$ and $- ε / 2$ lie in them at distance $ε / 2$ from $0$ (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 $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

Absolute value: $∣ 0 ∣ = 0$ ; $∣ u ∣ = 0$ exactly when $u = 0$ (Basic properties of the absolute value).

[L4]

Order in $R$ : trichotomy, so every real either equals $0$ or does not, exclusively; $0 < 1$ , so $1 \neq 0$ , and $ε / 2 > 0$ for $ε > 0$ (The multiplicative identity is positive, Ordered field).

[L5]

One-sided limits are the limits of the restrictions to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ (The left and right limits of $f$ at $c$ , as limits of the restrictions of $f$ to $A \cap (- \infty, c)$ and $A \cap (c, \infty)$ ).

[L6]

Locality: if two functions on $A$ agree on $A \cap N_{η}^{*} (c)$ for some real $η > 0$ , they have the same limits at $c$ (claim 1 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).

Counterexample

technique · direct

1.1

$f$ is a well-defined function on $R$ , by trichotomy; and $0 \in R$ is a limit point of $R$ .

L2L4

1.2

The reals $0$ and $1$ are distinct.

2.1

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

step 1.1L1L3L4

2.2

Both one-sided limits of $f$ at $0$ exist and equal $0$ : the point $0$ is a limit point of $(0, \infty)$ and of $(- \infty, 0)$ by [L2], and every $x$ in either set satisfies $x \neq 0$ , hence $f (x) = 0$ ; so any $δ > 0$ serves in the definition of each one-sided limit.

step 1.1L1L2L3L5

3.1

Yet $f (0) = 1$ , and $1 \neq 0 = \lim_{x \to 0} f (x)$ : at the point $c = 0$ of the domain, which is a limit point of the domain, the limit exists and differs from the value, refuting the claim.

step 1.2step 2.1L4

3.2

Changing the single value repairs the equality: $f$ and the constant function $g$ agree at every $x \neq 0$ , hence on $R \cap N_{1}^{*} (0)$ , so by [L6] the limit of $g$ at $0$ exists and equals $0$ ; and $g (0) = 0$ is that limit.

step 2.1L6

4.1

So the limit at a point of the domain is independent of the value of the function there, and the two agree only under an extra hypothesis on the function, never as a consequence of the limit existing.

step 3.1step 3.2∎

Remarks

A removable defect, not a jump. By step 2.2 the two one-sided limits exist and agree with each other and with the two-sided limit; the only disagreement is with the value. Compare the sign function (The sign function has both one-sided limits at $0$ and no two-sided limit), where the two one-sided limits exist and disagree, and no redefinition of the value can repair anything.
Why the witness is used again for composition. Because $f (0) \neq \lim_{y \to 0} f (y)$ , 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 $c$ fails for $f$ at $L = 0$ ; feeding it an inner function that takes the value $0$ then breaks the composition, which is With $g \equiv 0$ and $f$ equal to $0$ off the origin and $1$ at it, $\lim g = 0$ and $\lim_{y \to 0} f = 0$ while $f \circ g \equiv 1$ .
Nothing here depends on the particular values $0$ and $1$ , only on their being distinct. Any function constant off $c$ with a different value at $c$ refutes the claim in the same three lines.

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

With $g \equiv 0$ and $f$ equal to $0$ off the origin and $1$ at it, $\lim g = 0$ and $\lim_{y \to 0} f = 0$ while $f \circ g \equiv 1$

Statement refuted

Refuted claim: if $\lim_{x \to c} g (x) = L$ and $\lim_{y \to L} f (y) = M$ then the limit of $f \circ g$ at $c$ exists and equals $M$ — the false statement FALSE: $\lim_{x \to c} f (g (x)) = M$ whenever $\lim_{x \to c} g = L$ and $\lim_{y \to L} f = M$ .

Take $A = B = R$ , $c = 0$ , the constant function $g \equiv 0$ , and the function $f$ of The function equal to $0$ off the origin and to $1$ at the origin has limit $0 \neq 1$ there, equal to $0$ off the origin and to $1$ at it. Then $L = 0$ , $M = 0$ , and $f \circ g$ is the constant function $1$ , so the limit of $f \circ g$ at $0$ exists and equals $1 \neq 0 = M$ .

What this item adds to the false statement. It carries the comparison through: it identifies which of 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 $c$ fails here — both do — and it shows that replacing the inner function by the identity, which satisfies hypothesis (ii), restores the conclusion with the same outer function. So neither the outer function nor the composition operation is at fault; the failure is precisely that the inner function takes the critical value.

Facts & Assumptions

Given: The function $f : R \to R$ of The function equal to $0$ off the origin and to $1$ at the origin has limit $0 \neq 1$ there, with $f (y) = 0$ for $y \neq 0$ and $f (0) = 1$ ; the constant function $g : R \to R$ , $g (x) = 0$ ; the identity function $ι : R \to R$ , $ι (x) = x$ ; and the point $c : = 0$ .

[L1]

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

[L2]

[L3]

The witness function: $f (0) = 1$ by its definition, and the limit of $f$ at $0$ exists and equals $0$ , as verified in The function equal to $0$ off the origin and to $1$ at the origin has limit $0 \neq 1$ there.

[L4]

Absolute value: $∣ 0 ∣ = 0$ ; $∣ u ∣ = 0$ exactly when $u = 0$ (Basic properties of the absolute value).

[L5]

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

[L6]

Composition of limits, and its two extra hypotheses: (i) $L \in B$ and $f (L) = M$ ; (ii) some real $η > 0$ has $g (x) \neq L$ for every $x \in A$ with $0 < ∣ x - c ∣ < η$ (Composition of limits holds under either hypothesis: $f$ is defined at $L$ with value $M$ , or $g$ avoids $L$ on a punctured neighbourhood of $c$ ).

Counterexample

technique · direct

1.1

By [L3] the limit of $f$ at $0$ exists and equals $0$ , and $f (0) = 1$ ; so the outer hypothesis of the refuted claim holds with $L = 0$ and $M = 0$ .

1.2

$0$ is a limit point of $R$ , and $g (R) = {0} \subseteq R$ and $ι (R) = R$ , so both $f \circ g$ and $f \circ ι$ are functions on $R$ .

1.3

The reals $0$ and $1$ are distinct.

2.1

The inner hypothesis holds for $g$ with $L = 0$ : for every real $ε > 0$ every $δ > 0$ serves, since $∣ g (x) - 0 ∣ = ∣ 0 ∣ = 0 < ε$ for every $x$ . So the limit of $g$ at $0$ exists and equals $0$ .

step 1.2L1L4

2.2

It holds for $ι$ as well: given a real $ε > 0$ take $δ : = ε$ ; then $0 < ∣ x - 0 ∣ < δ$ gives $∣ ι (x) - 0 ∣ = ∣ x ∣ < ε$ . So the limit of $ι$ at $0$ exists and equals $0$ .

step 1.2L1

3.1

$f \circ g$ is the constant function $1$ : for every $x \in R$ , $g (x) = 0$ and hence $f (g (x)) = f (0) = 1$ . By the computation of step 2.1, applied to the constant $1$ in place of the constant $0$ , the limit of $f \circ g$ at $0$ exists and equals $1$ .

step 1.1step 2.1L1L4

3.2

$f \circ ι = f$ , since $f (ι (x)) = f (x)$ for every $x$ ; so by [L3] the limit of $f \circ ι$ at $0$ exists and equals $0 = M$ .

step 1.1step 2.2L3

4.1

Hence $\lim_{x \to 0} g (x) = 0 = L$ and $\lim_{y \to 0} f (y) = 0 = M$ , while $\lim_{x \to 0} f (g (x)) = 1 \neq 0 = M$ : the refuted claim is false.

step 1.3step 3.1L5

4.2

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 the pair $(f, g)$ : hypothesis (i) fails because $L = 0$ lies in $B = R$ while $f (L) = f (0) = 1 \neq 0 = M$ , and hypothesis (ii) fails because $g (x) = 0 = L$ for every $x$ , so no punctured neighbourhood of $0$ avoids the value $L$ . For the pair $(f, ι)$ , hypothesis (ii) does hold with $η : = 1$ , since $ι (x) = x \neq 0$ whenever $0 < ∣ x - 0 ∣ < 1$ ; and step 3.2 confirms the conclusion of the theorem there.

step 1.1step 3.1step 3.2L6

5.1

So the two safeguards in the true theorem cannot both be omitted, and the obstruction is located exactly at the values of the inner function that equal $L$ .

step 4.1step 4.2∎

Remarks

The same outer function serves both roles. With $g$ the composition fails, with $ι$ it succeeds, and $f$ is unchanged. So the failure cannot be attributed to any pathology of $f$ beyond the one recorded in The function equal to $0$ off the origin and to $1$ at the origin has limit $0 \neq 1$ there: that its value at $0$ differs from its limit at $0$ .
Constancy of $g$ is not the issue either. What matters is that $g$ takes the value $L$ on every punctured neighbourhood of $c$ . Any inner function doing that, constant or not, produces the same failure by the same argument, since the outer estimate is unavailable at those arguments.
The practical rule. When substituting $y = g (x)$ inside a limit, check one of 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 $c$ : either the outer function is defined at $L$ with the right value there, or the inner function avoids $L$ near $c$ . Substitutions such as $y = 1 / x$ satisfy the second for structural reasons; substitutions into a function known only through its limit satisfy neither in general.

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

The indicator of $Q$ has a limit at no point of $R$

Statement refuted

Write $Q$ for the canonical copy of the rationals inside $R$ (The rationals embed densely in the reals), $X : = R ∖ Q$ for the irrationals, and let

$1_{Q} : R \to R, 1_{Q} (x) : = {\begin{cases} 1 & x \in Q, \\ 0 & x \in X . \end{cases}$

Refuted claim: there is a point $c \in R$ at which $1_{Q}$ has a limit (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ).

The refutation fixes an arbitrary real $c$ and produces two sequences tending to $c$ , one of rationals and one of irrationals, both avoiding $c$ ; the image sequences are constantly $1$ and constantly $0$ , and A function has no limit at $c$ as soon as two sequences in $A ∖ {c}$ tending to $c$ give different limits of the values applies. Since $c$ was arbitrary, the function has a limit nowhere.

Where the choice principle enters, and where it does not. Producing the two sequences is a use of A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed, whose left-to-right direction spends countable choice, and that cost is inherited here and recorded by that item. The criterion applied afterwards is the choice-free one (A function has no limit at $c$ as soon as two sequences in $A ∖ {c}$ tending to $c$ give different limits of the values).

Facts & Assumptions

Given: The canonical copy $Q \subseteq R$ of the rationals, the irrationals $X = R ∖ Q$ , the function $1_{Q}$ above, and an arbitrary real $c$ .

[L1]

Density: $Q$ and $X$ are both dense in $R$ , that is, each has closure $R$ (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable); and the closure of a set $S$ is exactly the set of points every neighbourhood of which meets $S$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $R$ , Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L2]

Sequential characterisation of the closure: $x$ lies in the closure of $S$ if and only if there is a sequence with all terms in $S$ converging to $x$ (A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). The direction used below, from the closure to a sequence, is the one that spends countable choice, as that item records.

[L3]

Neighbourhoods: $N_{ρ} (u) = {y : ∣ y - u ∣ < ρ}$ for real $ρ > 0$ , so $N_{ε / 2} (c + ε / 2) = {y : c < y < c + ε}$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L4]

Nonexistence criterion: if two sequences with all terms in $A ∖ {c}$ converge to $c$ while the image sequences converge to distinct reals, then the function has no limit at $c$ (A function has no limit at $c$ as soon as two sequences in $A ∖ {c}$ tending to $c$ give different limits of the values).

[L5]

[L6]

A constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L7]

Absolute value and order: $∣ u ∣ \geq 0$ and $∣ u ∣ = u$ for $u \geq 0$ (Basic properties of the absolute value); $0 < 1$ , so $2 > 0$ , $ε / 2 > 0$ and $ε / 2 < ε$ for $ε > 0$ ; trichotomy and totality (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Ordered field).

Counterexample

technique · direct

1.1

Let $c \in R$ be arbitrary. Then $c$ is a limit point of $R$ , the domain of $1_{Q}$ , so the question of a limit at $c$ is well posed.

1.2

Let $S$ be either $Q$ or $X$ , and let $ε > 0$ be an arbitrary real. Applying [L1] at the real $c + ε / 2$ with the radius $ε / 2 > 0$ , the neighbourhood $N_{ε / 2} (c + ε / 2)$ meets $S$ ; and by [L3] every $y$ in that neighbourhood satisfies $c < y < c + ε$ , hence $y \neq c$ and $0 < ∣ y - c ∣ < ε$ . So every neighbourhood of $c$ meets $S ∖ {c}$ .

L1L3L7

2.1

By [L1] again, step 1.2 says exactly that $c$ lies in the closure of $Q ∖ {c}$ and in the closure of $X ∖ {c}$ . Hence [L2] supplies a sequence $(q_{k})$ with all terms in $Q ∖ {c}$ converging to $c$ , and a sequence $(u_{k})$ with all terms in $X ∖ {c}$ converging to $c$ .

step 1.2L1L2choose

3.1

Every term of $(q_{k})$ lies in $Q$ , so $1_{Q} (q_{k}) = 1$ for every $k$ and the image sequence is the constant sequence $1$ , converging to $1$ ; every term of $(u_{k})$ lies in $X$ , so $1_{Q} (u_{k}) = 0$ for every $k$ and that image sequence converges to $0$ . The reals $1$ and $0$ are distinct.

step 2.1L6L7

4.1

Both sequences have all their terms in $R ∖ {c}$ and converge to $c$ , while their image sequences converge to distinct reals; by [L4] the function $1_{Q}$ has no limit at $c$ . Since $c \in R$ was arbitrary, it has a limit at no point of $R$ .

step 1.1step 2.1step 3.1L4∎

Remarks

Why the sets are punctured before the sequences are drawn. A function has no limit at $c$ as soon as two sequences in $A ∖ {c}$ tending to $c$ give different limits of the values requires all terms to lie in $A ∖ {c}$ , since a sequence allowed to take the value $c$ would report on $1_{Q} (c)$ , which the limit ignores (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ). Step 1.2 therefore verifies density of $Q ∖ {c}$ and of $X ∖ {c}$ directly, by placing the auxiliary neighbourhood strictly to the right of $c$ .
The two densities are not proved the same way. $Q$ is dense because it is built to approximate; $X$ is dense because a countable set cannot exhaust an interval. Both are claims 1 and 2 of Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, and this item uses them only through the neighbourhood formulation of [L1].
The failure is as total as possible. Not merely does the limit fail at some points: it fails at every point of $R$ , while the function is bounded throughout, taking only the values $0$ and $1$ . Multiplying by $x$ leaves exactly one point where a limit survives, which is $x \cdot 1_{Q} (x)$ has a limit at $0$ and at no other point.

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

$x \cdot 1_{Q} (x)$ has a limit at $0$ and at no other point

Example

With $1_{Q}$ as in The indicator of $Q$ has a limit at no point of $R$ , let

$d : R \to R, d (x) : = x \cdot 1_{Q} (x),$

so $d (x) = x$ for rational $x$ and $d (x) = 0$ for irrational $x$ . Then the limit of $d$ at $0$ exists, with

$\lim_{x \to 0} d (x) = 0,$

and at every $c \neq 0$ the function $d$ has no limit.

The point of the example. The factor $1_{Q}$ has a limit nowhere; multiplying it by $x$ repairs exactly one point, and only that one. The repair at $0$ is the squeeze theorem (If $f \leq g \leq h$ near $c$ and $f$ and $h$ have the same limit at $c$ , then so does $g$ ) applied to $- ∣ x ∣ \leq d (x) \leq ∣ x ∣$ ; the failure elsewhere is the same two-sequence argument as in The indicator of $Q$ has a limit at no point of $R$ , now with image limits $c$ and $0$ , which are distinct precisely because $c \neq 0$ .

Facts & Assumptions

Given: The canonical copy $Q \subseteq R$ of the rationals, the irrationals $X = R ∖ Q$ , the function $d (x) = x \cdot 1_{Q} (x)$ , and a real $c \neq 0$ .

[L1]

The values of $d$ : $d (x) = x$ for $x \in Q$ and $d (x) = 0$ for $x \in X$ ; every real lies in exactly one of $Q$ and $X$ (The indicator of $Q$ has a limit at no point of $R$ ).

[L2]

[L3]

Squeeze theorem: if $f \leq g \leq k$ on $A \cap N_{η}^{*} (c)$ for some real $η > 0$ and the limits of $f$ and of $k$ at $c$ exist and are equal to $L$ , then the limit of $g$ at $c$ exists and equals $L$ (If $f \leq g \leq h$ near $c$ and $f$ and $h$ have the same limit at $c$ , then so does $g$ ).

[L4]

Nonexistence criterion (A function has no limit at $c$ as soon as two sequences in $A ∖ {c}$ tending to $c$ give different limits of the values).

[L5]

For every real $c$ there are a sequence $(q_{k})$ with all terms in $Q ∖ {c}$ and a sequence $(u_{k})$ with all terms in $X ∖ {c}$ , both converging to $c$ . This is exactly what is established in the course of The indicator of $Q$ has a limit at no point of $R$ , from density (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $R$ , The rationals embed densely in the reals) and the sequential characterisation of the closure (A point lies in the closure of $A \subseteq R$ iff some sequence in $A$ converges to it, so a subset of $R$ is closed iff it is sequentially closed, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals); the countable choice spent there is inherited here.

[L6]

[L7]

Absolute value: $∣ u ∣ \geq 0$ ; $- ∣ u ∣ \leq u \leq ∣ u ∣$ ; $∣ u ∣ = u$ for $u \geq 0$ ; $∣ 0 ∣ = 0$ (Basic properties of the absolute value). Order arithmetic: trichotomy and totality; $0 < 1$ , so $2 > 0$ and $ε / 2 < ε$ for $ε > 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field).

[L8]

A constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Verification

technique · direct

1.1

For every $x \in R$ , $- ∣ x ∣ \leq d (x) \leq ∣ x ∣$ : if $x \in Q$ then $d (x) = x$ and $- ∣ x ∣ \leq x \leq ∣ x ∣$ ; if $x \in X$ then $d (x) = 0$ and $- ∣ x ∣ \leq 0 \leq ∣ x ∣$ .

L1L7

1.2

Every real is a limit point of $R$ ; in particular $0$ and the given $c$ are.

1.3

The functions $x \mapsto - ∣ x ∣$ and $x \mapsto ∣ x ∣$ have limit $0$ at $0$ : given a real $ε > 0$ take $δ : = ε$ ; every $x$ with $0 < ∣ x - 0 ∣ < δ$ satisfies $∣ ∣ x ∣ - 0 ∣ = ∣ x ∣ < ε$ and $∣ - ∣ x ∣ - 0 ∣ = ∣ x ∣ < ε$ .

L2L7

2.1

The three functions satisfy $- ∣ x ∣ \leq d (x) \leq ∣ x ∣$ on all of $R$ , in particular on $R \cap N_{1}^{*} (0)$ , and the outer two have limit $0$ at $0$ ; since $0$ is a limit point of $R$ , the squeeze theorem [L3] gives that the limit of $d$ at $0$ exists and equals $0$ .

step 1.1step 1.2step 1.3L3

2.2

Fix the real $c \neq 0$ . By [L5] there are a sequence $(q_{k})$ with all terms in $Q ∖ {c}$ and a sequence $(u_{k})$ with all terms in $X ∖ {c}$ , both converging to $c$ .

step 1.2L5choose

3.1

By [L1], $d (q_{k}) = q_{k}$ for every $k$ , so the image sequence $(d (q_{k}))$ is $(q_{k})$ itself and converges to $c$ ; and $d (u_{k}) = 0$ for every $k$ , so that image sequence is constant and converges to $0$ . Since $c \neq 0$ , the two limits are distinct, and both sequences have all their terms in $R ∖ {c}$ and converge to $c$ ; by [L4] the function $d$ has no limit at $c$ .

step 2.2L1L4L7L8

4.1

So the limit of $d$ exists at $0$ , with value $0$ , and fails to exist at every other real: $d$ has a limit at exactly one point.

step 2.1step 3.1∎

Remarks

Why $0$ is the exceptional point. The squeeze bound $∣ d (x) ∣ \leq ∣ x ∣$ is useful only where $∣ x ∣$ is small, that is near $0$ ; at any other $c$ the two bounding functions have limit $∣ c ∣ \neq 0$ and $- ∣ c ∣$ , which are different, so the squeeze theorem says nothing there. That is not an accident of the proof: the two-sequence argument shows the limit genuinely fails at every such $c$ .
The value $d (0) = 0$ happens to equal the limit, since $0$ is rational, so $d$ satisfies at $0$ the equality that FALSE: $\lim_{x \to c} f (x) = f (c)$ whenever both sides exist shows is not automatic. It is the only point of $R$ at which $d$ does so.
Contrast with $ψ (1 / x)$ . There the oscillation is bounded and the failure is confined to a single point, $0$ , with the multiplication by $x$ repairing precisely that point ( $x ψ (1 / x) \to 0$ as $x \to 0$ , by the squeeze theorem). Here the failure is everywhere and the multiplication repairs precisely one point. The two examples are the same mechanism — a bounded factor damped by a vanishing one — applied to opposite kinds of irregularity.

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

On the domain ${0} \cup [1, 2]$ every real is vacuously a limit at $0$

Statement refuted

Refuted claim: for every $A \subseteq R$ , every $f : A \to R$ and every $c \in A$ , at most one real $L$ satisfies

$(\forall ε > 0) (\exists δ > 0) (\forall x \in A) [0 < ∣ x - c ∣ < δ ⟹ ∣ f (x) - L ∣ < ε]$

— the false statement FALSE: a function has at most one limit at every point of its domain, isolated points included.

The witness is $A : = {0} \cup [1, 2]$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), $f : A \to R$ the constant $0$ , and $c : = 0$ . At $c$ the displayed formula holds for every real $L$ at once, so it determines nothing.

What this item adds. It exhibits the dichotomy inside one example: at the isolated point $0$ the formula is vacuous, while at the point $1$ of the same domain — which is a limit point of $A$ — the formula is not vacuous and At a limit point of the domain a function has at most one limit applies, so the limit there exists and is unique. The same $A$ , the same $f$ , and opposite behaviour at two of its points.

Facts & Assumptions

Given: The set $A : = {0} \cup [1, 2]$ , the constant function $f : A \to R$ with $f (x) : = 0$ for every $x \in A$ , and the points $0$ and $1$ of $A$ .

[L1]

The $ε$ - $δ$ formula displayed above, and the fact that The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ imposes it only at a limit point of the domain, where At a limit point of the domain a function has at most one limit then makes $L$ unique.

[L2]

Limit point and isolated point: $c$ is a limit point of $S$ when $N_{ρ}^{*} (c) \cap S \neq \emptyset$ for every real $ρ > 0$ ; $c \in S$ is isolated in $S$ when $N_{ρ} (c) \cap S = {c}$ for some real $ρ > 0$ ; and for $c \in S$ the two 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: $N_{ρ} (u) = {y : ∣ y - u ∣ < ρ}$ and $N_{ρ}^{*} (u) = N_{ρ} (u) ∖ {u}$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L4]

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

[L5]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ exactly when $u = 0$ ; $∣ u ∣ = u$ for $u \geq 0$ ; $∣ 0 ∣ = 0$ (Basic properties of the absolute value).

[L6]

Order in $R$ : trichotomy and totality; $0 < 1$ , so $2 > 0$ and $ρ / 2 > 0$ with $ρ / 2 < ρ$ for $ρ > 0$ ; and of two positive reals the smaller is positive (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field).

Counterexample

technique · direct

1.1

$A = {0} \cup [1, 2]$ is a subset of $R$ , and $f$ is the constant $0$ on $A$ ; both $0$ and $1$ belong to $A$ .

1.2

$0$ is an isolated point of $A$ and not a limit point of $A$ : $N_{1} (0) \cap A = {0}$ , since an element of $A$ is either $0$ , with $∣ 0 - 0 ∣ = 0 < 1$ , or an element of $[1, 2]$ , with $∣ y - 0 ∣ = y \geq 1$ and hence outside $N_{1} (0)$ .

L2L3L4L5

1.3

The reals $0$ and $1$ are distinct.

2.1

Take $δ : = 1$ . No $x \in A$ satisfies $0 < ∣ x - 0 ∣ < 1$ : such an $x$ would lie in $N_{1}^{*} (0) \cap A$ , which is contained in $N_{1} (0) \cap A = {0}$ and excludes $0$ , hence is empty. So for every real $L$ and every real $ε > 0$ the choice $δ = 1$ makes the implication vacuously true, and every real $L$ satisfies the displayed formula at $c = 0$ .

step 1.2L1L3L5

2.2

By contrast $1 \in A$ is a limit point of $A$ : given a real $ρ > 0$ , let $σ$ be the smaller of $ρ$ and $1$ , so $σ > 0$ ; then $1 + σ / 2$ satisfies $1 \leq 1 + σ / 2 \leq 1 + 1 / 2 \leq 2$ , so it lies in $[1, 2] \subseteq A$ , and $0 < ∣ (1 + σ / 2) - 1 ∣ = σ / 2 < ρ$ . There The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ applies, At a limit point of the domain a function has at most one limit gives at most one $L$ , and in fact $\lim_{x \to 1} f (x) = 0$ , since $∣ f (x) - 0 ∣ = ∣ 0 ∣ = 0 < ε$ for every $x \in A$ and every real $ε > 0$ .

step 1.1L1L2L4L5L6

3.1

In particular $L = 0$ and $L = 1$ both satisfy the formula at $c = 0$ , and they are distinct: more than one real satisfies it, so the claim is refuted. This is why The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ is stated only at a limit point, and why $\lim_{x \to 0} f (x)$ is left undefined on this domain.

step 1.3step 2.1L1L6

4.1

So on one and the same domain the formula pins down a unique value at the limit point $1$ and no value at all at the isolated point $0$ : uniqueness of the limit is a property of limit points, not of arbitrary points of the domain.

step 2.2step 3.1∎

Remarks

Nothing about $f$ is used at $0$ . Step 2.1 never evaluates the function: the implication has no instances. Any $f$ whatever on this $A$ would give the same conclusion, which is precisely why the formula carries no information there.
The alternatives are exact. At a limit point of the domain the formula has at most one solution (At a limit point of the domain a function has at most one limit), while at an isolated point every real solves it (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).
Some texts declare the limit at an isolated point to be $f (c)$ . That convention is consistent — it selects one of the many solutions — but it is a stipulation, not a theorem, and FALSE: a function has at most one limit at every point of its domain, isolated points included records why this library declines it.

Limits of Real Functions: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The classical form of the oscillator above is sin⁡(1/x)sin(1/x), which this library can only construct much later

Orientation, not a claim of this library

The later analytic construction

What $\psi$ supplies instead

What is genuinely lost, and what is not

5 · Examples, counterexamples and false statements

Every polynomial has lim⁡x→cp(x)=p(c)limx→c​p(x)=p(c), and rational functions do so away from the zeros of the denominator

Example

Facts & Assumptions

Verification

Remarks

The trigonometry-free oscillator ψ(x)=inf⁡n∈Z∣x−n∣ψ(x)=infn∈Z​∣x−n∣ is well defined and attained at a nearest integer, takes values in [0,1/2][0,1/2], vanishes exactly on ZZ, equals 1/21/2 at half-integers, and is 11-periodic

Example

Facts & Assumptions

Verification

Remarks

x ψ(1/x)→0xψ(1/x)→0 as x→0x→0, by the squeeze theorem

Example

Facts & Assumptions

Verification

Remarks

The sign function has both one-sided limits at 00 and no two-sided limit

Example

Facts & Assumptions

Verification

Remarks

(3x2−1)/(x2+x)→3(3x2−1)/(x2+x)→3 as x→+∞x→+∞

Example

Facts & Assumptions

Verification

Remarks

ψ(1/x)ψ(1/x) has no limit at 00: two sequences tending to 00 give values constantly 00 and constantly 1/21/2

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The function equal to 00 off the origin and to 11 at the origin has limit 0≠10=1 there

Statement refuted

Facts & Assumptions

Counterexample

Remarks

With g≡0g≡0 and ff equal to 00 off the origin and 11 at it, lim⁡g=0limg=0 and lim⁡y→0f=0limy→0​f=0 while f∘g≡1f∘g≡1

Statement refuted

Facts & Assumptions

Counterexample

Remarks

The indicator of QQ has a limit at no point of RR

Statement refuted

Facts & Assumptions

Counterexample

Remarks

x⋅1Q(x)x⋅1Q​(x) has a limit at 00 and at no other point

Example

Facts & Assumptions

Verification

Remarks

On the domain {0}∪[1,2]{0}∪[1,2] every real is vacuously a limit at 00

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

The classical form of the oscillator above is $\sin (1 / x)$ , which this library can only construct much later

Every polynomial has $\lim_{x \to c} p (x) = p (c)$ , and rational functions do so away from the zeros of the denominator

The trigonometry-free oscillator $ψ (x) = \inf_{n \in Z} ∣ x - n ∣$ is well defined and attained at a nearest integer, takes values in $[0, 1 / 2]$ , vanishes exactly on $Z$ , equals $1 / 2$ at half-integers, and is $1$ -periodic

$x ψ (1 / x) \to 0$ as $x \to 0$ , by the squeeze theorem

The sign function has both one-sided limits at $0$ and no two-sided limit

$(3 x^{2} - 1) / (x^{2} + x) \to 3$ as $x \to + \infty$

$ψ (1 / x)$ has no limit at $0$ : two sequences tending to $0$ give values constantly $0$ and constantly $1 / 2$

The function equal to $0$ off the origin and to $1$ at the origin has limit $0 \neq 1$ there

With $g \equiv 0$ and $f$ equal to $0$ off the origin and $1$ at it, $\lim g = 0$ and $\lim_{y \to 0} f = 0$ while $f \circ g \equiv 1$

The indicator of $Q$ has a limit at no point of $R$

$x \cdot 1_{Q} (x)$ has a limit at $0$ and at no other point

On the domain ${0} \cup [1, 2]$ every real is vacuously a limit at $0$