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

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

Statement refuted

Refuted claim: the function

$\Phi : \mathbb{R} \setminus \{0\} \to \mathbb{R}, \qquad \Phi(x) := \psi(1/x) ,$

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

$\Phi$ is bounded — $0 \le \Phi(x) \le 1/2$ for every $x \ne 0$ , by claim 2 of The trigonometry-free oscillator $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ is well defined and attained at a nearest integer, takes values in $[0, 1/2]$ , vanishes exactly on $\mathbb{Z}$ , 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 $\Phi$ is constantly $0$ and constantly $1/2$ , and applies A function has no limit at $c$ as soon as two sequences in $A \setminus \{c\}$ tending to $c$ give different limits of the values.

Facts & Assumptions

Given: The function $\Phi(x) = \psi(1/x)$ on $\mathbb{R} \setminus \{0\}$ , and the sequences $x_k := 1/(k+1)$ and $y_k := 2/(2k+1)$ for $k \in \mathbb{N}$ . Sequences are functions on $\mathbb{N}$ and $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, The natural numbers $\mathbb{N}$ (von Neumann)), so the first terms are $x_0 = 1$ and $y_0 = 2$ ; the denominators $k+1$ and $2k+1$ are canonical naturals $\ge 1$ , never $0$ , which is why the sequences are written this way and not as $1/k$ .

[L1]

The function $\psi$ vanishes exactly on $\mathbb{Z}$ , satisfies $\psi(m + 1/2) = 1/2$ for every $m \in \mathbb{Z}$ , and takes values in $[0,1/2]$ (The trigonometry-free oscillator $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ is well defined and attained at a nearest integer, takes values in $[0, 1/2]$ , vanishes exactly on $\mathbb{Z}$ , 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 $A \setminus \{c\}$ 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 $A \setminus \{c\}$ 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 $\varepsilon > 0$ there is a natural $n \ge 1$ with $1/n < \varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , 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 $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L6]

Absolute value (Basic properties of the absolute value); order and field arithmetic: $0 < 1$ , so $2 > 0$ and $1/2 > 0$ with $1/2 \ne 0$ (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 $\mathbb{R}$ : every canonical natural is an integer, and $\mathbb{Z}$ 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 $\mathbb{R} \setminus \{0\}$ : given a real $\varepsilon > 0$ , the real $\varepsilon/2$ is positive, hence lies in $\mathbb{R} \setminus \{0\}$ , and $0 < |\varepsilon/2 - 0| = \varepsilon/2 < \varepsilon$ .

L5L6

1.2

For every $k \in \mathbb{N}$ the terms $x_k = 1/(k+1)$ and $y_k = 2/(2k+1)$ are defined and positive, since $k + 1 \ge 1 > 0$ and $2k + 1 \ge 1 > 0$ ; in particular $x_k \ne 0$ and $y_k \ne 0$ , so both sequences have all their terms in $\mathbb{R} \setminus \{0\}$ , which equals $(\mathbb{R} \setminus \{0\}) \setminus \{0\}$ .

L4L6L7

1.3

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

2.1

$x_k \to 0$ : given a real $\varepsilon > 0$ , [L4] supplies a natural $n \ge 1$ with $1/n < \varepsilon$ ; every $k \ge n$ has $k + 1 > n \ge 1$ , hence $|x_k - 0| = 1/(k+1) < 1/n < \varepsilon$ .

step 1.2L3L4L6

2.2

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

step 1.2L3L4L6

2.3

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

step 1.2L1L3L7

2.4

$\Phi(y_k) = 1/2$ for every $k$ : $1/y_k = (2k+1)/2 = k + 1/2$ with $k$ an integer by [L7], so $\psi(1/y_k) = 1/2$ 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 $(\mathbb{R} \setminus \{0\}) \setminus \{0\}$ and both converge to $0$ , which is a limit point of $\mathbb{R} \setminus \{0\}$ , while the image sequences converge to the distinct reals $0$ and $1/2$ . By [L2], $\Phi$ 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 $\lim_{x \to 0^{+}} \Phi(x)$ (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 $2/(2k+1)$ . 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 $\psi(x) = \inf_{n \in \mathbb{Z}} |x - n|$ is well defined and attained at a nearest integer, takes values in $[0, 1/2]$ , vanishes exactly on $\mathbb{Z}$ , equals $1/2$ at half-integers, and is $1$ -periodic evaluate $\psi$ exactly. Writing $1/k$ instead would be undefined at the index $k = 0$ , since $\mathbb{N}$ contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).
Multiplying by $x$ repairs it. The function $x \mapsto x\,\psi(1/x)$ does have a limit at $0$ , namely $0$ , by the squeeze theorem ( $x\,\psi(1/x) \to 0$ 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 $\psi(1/x)$ ; 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.

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Sources

Limit of a function (Wikipedia) (standard reference, not scraped)
Classification of discontinuities (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)

ψ(1/x)\psi(1/x)ψ(1/x) has no limit at 000: two sequences tending to 000 give values constantly 000 and constantly 1/21/21/2