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

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If $f \leq g$ on a punctured neighbourhood of $c$ then $\lim f \leq \lim g$ , non-strictly

Statement

Let $A \subseteq R$ , let $c$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ), let $f, g : A \to R$ and suppose both limits at $c$ exist (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ). Suppose further that there is a real $η > 0$ with

$f (x) \leq g (x) for every x \in A with 0 < ∣ x - c ∣ < η .$

Then

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

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

Only the values near $c$ matter, by The limit at $c$ depends only on the restriction of $f$ to a punctured neighbourhood of $c$ , and passes to any subset of the domain having $c$ as a limit point: the hypothesis is imposed on a punctured neighbourhood of $c$ and on nothing else, and it says nothing about $f (c)$ and $g (c)$ , which the definition ignores in any case.

Facts & Assumptions

Given: A set $A \subseteq R$ , a limit point $c$ of $A$ , functions $f, g : A \to R$ , reals $L, M$ with $\lim_{x \to c} f (x) = L$ and $\lim_{x \to c} g (x) = M$ , and a real $η > 0$ with $f (x) \leq g (x)$ for every $x \in A$ satisfying $0 < ∣ x - c ∣ < η$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ).

[L1]

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

[L2]

Limit point: for every real $ρ > 0$ there is $x \in A$ with $0 < ∣ x - c ∣ < ρ$ (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

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

[L4]

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

Proof

technique · contradiction

1.1

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

assume-contra

2.1

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

step 1.1L4

3.1

By [L1] fix reals $δ_{1}, δ_{2} > 0$ such that every $x \in A$ with $0 < ∣ x - c ∣ < δ_{1}$ has $∣ f (x) - L ∣ < ε$ and every $x \in A$ with $0 < ∣ x - c ∣ < δ_{2}$ has $∣ g (x) - M ∣ < ε$ ; let $δ$ be the smallest of $δ_{1}$ , $δ_{2}$ and $η$ , so $δ > 0$ .

step 2.1L1L4choose

4.1

Since $c$ is a limit point of $A$ , fix $x \in A$ with $0 < ∣ x - c ∣ < δ$ .

step 3.1L2choose

5.1

That $x$ satisfies $0 < ∣ x - c ∣ < δ_{1}$ and $0 < ∣ x - c ∣ < δ_{2}$ , so $∣ f (x) - L ∣ < ε$ gives $f (x) > L - ε$ and $∣ g (x) - M ∣ < ε$ gives $g (x) < M + ε$ ; since $L - ε = M + ε$ , this yields $g (x) < f (x)$ .

step 3.1step 4.1L3L4

6.1

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

step 3.1step 5.1L4

7.1

The assumption that $L \leq M$ fails is therefore untenable, and $\lim_{x \to c} f (x) = L \leq M = \lim_{x \to c} g (x)$ .

step 6.1L4discharge-contradiction∎

Remarks

Both limits are assumed to exist. Nothing here proves existence: the statement compares two numbers that are given. The theorem that produces a limit from an order hypothesis 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 $g$ , whose conclusion is exactly the existence of the middle limit.
The special case $f \equiv 0$ says that a function which is non-negative near $c$ has a non-negative limit there. Its contrapositive is the form used in practice: a negative limit forces negative values near $c$ , which is a weak version of 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.
The sequential analogue is Limits preserve non-strict inequalities, and it too is non-strict for the same reason: the counterexample is a strict inequality between quantities whose difference tends to $0$ .

Depends on

Used by

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)

If f≤gf≤g on a punctured neighbourhood of cc then lim⁡f≤lim⁡glimf≤limg, non-strictly