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

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

Statement refuted

Refuted claim: if $c \in A$ is a limit point of $A \subseteq \mathbb{R}$ and $f : A \to \mathbb{R}$ 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

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

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

[L1]

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

[L2]

Limit point: every real is a limit point of $\mathbb{R}$ , punctured neighbourhoods being never empty; and $0$ is a limit point of $(0,\infty)$ and of $(-\infty,0)$ , since $\varepsilon/2$ and $-\varepsilon/2$ lie in them at distance $\varepsilon/2$ from $0$ (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 $\mathbb{R}$ , Intervals of $\mathbb{R}$ : 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 $\mathbb{R}$ : trichotomy, so every real either equals $0$ or does not, exclusively; $0 < 1$ , so $1 \ne 0$ , and $\varepsilon/2 > 0$ for $\varepsilon > 0$ (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 $A \cap N^{*}_{\eta}(c)$ for some real $\eta > 0$ , 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 $\mathbb{R}$ , by trichotomy; and $0 \in \mathbb{R}$ is a limit point of $\mathbb{R}$ .

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 $\varepsilon > 0$ , take $\delta := 1 > 0$ ; every $x \in \mathbb{R}$ with $0 < |x - 0| < 1$ has $|x| \ne 0$ , hence $x \ne 0$ , hence $f(x) = 0$ and $|f(x) - 0| = |0| = 0 < \varepsilon$ .

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 $x \ne 0$ , hence $f(x) = 0$ ; so any $\delta > 0$ serves in the definition of each one-sided limit.

step 1.1L1L2L3L5

3.1

Yet $f(0) = 1$ , and $1 \ne 0 = \lim_{x \to 0} f(x)$ : 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 $x \ne 0$ , hence on $\mathbb{R} \cap N^{*}_{1}(0)$ , 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 $f(0) \ne \lim_{y \to 0} f(y)$ , 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.

Depends on

Used by

With g ≡ 0 and f equal to 0 off the origin and 1 at it, lim g = 0 and lim_y → 0 f = 0 while f ∘ g ≡ 1 Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, §3.1: Limits of functions (standard reference, not scraped)
Classification of discontinuities (Wikipedia) (standard reference, not scraped)
Limit of a function (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.1 (standard reference, not scraped)

The function equal to 000 off the origin and to 111 at the origin has limit 0≠10 \ne 10=1 there