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

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FALSE: $f < g$ near $c$ implies $\lim f < \lim g$

Statement

False claim: let $A \subseteq \mathbb{R}$ , let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ), let $f, g : A \to \mathbb{R}$ have limits at $c$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ), and suppose there is a real $\eta > 0$ with

$f(x) < g(x) \qquad \text{for every } x \in A \text{ with } 0 < |x - c| < \eta .$

Then $\lim_{x \to c} f(x) < \lim_{x \to c} g(x)$ .

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

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

Facts & Assumptions

Given: The set $A := \mathbb{R}$ , the point $c := 0$ , the constant function $f : \mathbb{R} \to \mathbb{R}$ with $f(x) := 0$ for every $x$ , and the function $g : \mathbb{R} \to \mathbb{R}$ with $g(x) := |x|$ .

[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 of $h$ with $0 < |x - c| < \delta$ satisfies $|h(x) - P| < \varepsilon$ .

[L2]

Every real is a limit point of $\mathbb{R}$ , punctured neighbourhoods being never empty (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ , The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L3]

Absolute value: $|u| \ge 0$ ; $|u| = 0$ exactly when $u = 0$ ; and $|u| = u$ for $u \ge 0$ , so $\bigl| |x| - 0 \bigr| = |x|$ (Basic properties of the absolute value).

[L4]

Order in $\mathbb{R}$ : trichotomy, so $|u| \ge 0$ together with $|u| \ne 0$ gives $|u| > 0$ , and $t < t$ is impossible (Ordered field).

Refutation

technique · direct

1.1

The point $0$ is a limit point of $\mathbb{R}$ .

1.2

The strict hypothesis holds with $\eta := 1$ : every $x \in \mathbb{R}$ with $0 < |x - 0| < 1$ has $|x| \ne 0$ , hence $|x| > 0 = f(x)$ , that is $f(x) < g(x)$ .

L3L4

2.1

Both limits exist and are equal to $0$ . For $f$ : $|f(x) - 0| = 0 < \varepsilon$ for every $x$ and every real $\varepsilon > 0$ , any $\delta$ serving. For $g$ : given a real $\varepsilon > 0$ take $\delta := \varepsilon$ ; every $x$ with $0 < |x - 0| < \delta$ satisfies $\bigl| |x| - 0 \bigr| = |x| < \varepsilon$ .

step 1.1L1L3

3.1

So $f(x) < g(x)$ throughout a punctured neighbourhood of $0$ while $\lim_{x \to 0} f(x) = 0 = \lim_{x \to 0} g(x)$ ; the asserted strict inequality $0 < 0$ is impossible by trichotomy, so the claim is false.

step 1.2step 2.1L4∎

Remarks

The non-strict conclusion is sharp, and this witness shows it: the hypothesis is as strong as a pointwise strict inequality can be, and the conclusion still degenerates to equality.
The same phenomenon for sequences is the reason Limits preserve non-strict inequalities is stated non-strictly; the witness there is a positive null sequence compared with the constant $0$ , which is the sequential shadow of the pair above.
A common misuse. From $f < g$ near $c$ one may conclude $\lim f \le \lim g$ and nothing more; in particular one may not conclude that $\lim g > 0$ from $g > 0$ near $c$ . To get a strict conclusion one needs either a uniform gap, as noted above, or a separate argument such as If $\lim_{x \to c} f(x) = L \ne 0$ then $|f| > |L|/2$ on a punctured neighbourhood of $c$ ; in particular if $L > 0$ then $f > L/2 > 0$ there, which works in the opposite direction: from a nonzero limit to a bound on the values.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources

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

FALSE: f<gf < gf<g near ccc implies lim⁡f<lim⁡g\lim f < \lim glimf<limg