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

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If $f \leq g \leq h$ near $c$ and $f$ and $h$ have the same limit at $c$ , then so does $g$

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 $R$ ) and let $f, g, h : A \to R$ . Suppose there is a real $η > 0$ with

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

and suppose the limits of $f$ and of $h$ at $c$ exist and are equal, say $\lim_{x \to c} f (x) = \lim_{x \to c} h (x) = L$ (The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ ). Then the limit of $g$ at $c$ exists, and

$\lim_{x \to c} g (x) = \lim_{x \to c} f (x) = \lim_{x \to c} h (x) = L .$

This is the one result on this page that produces a limit rather than computing one. No hypothesis whatever is placed on $g$ beyond the two inequalities: $g$ may be wildly irregular, as $x \mapsto x ψ (1 / x)$ on the companion page is, and the theorem still delivers its limit at $c$ .

The proof is a direct $ε$ - $δ$ argument and uses no choice principle.

Facts & Assumptions

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

[L2]

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

[L3]

Order arithmetic in $R$ : the order is transitive, and mixed chains compose, so $u < v \leq w$ gives $u < w$ and $u \leq v < w$ gives $u < w$ ; adding a constant to an inequality (Order is preserved by adding a constant and by adding inequalities); of finitely many positive reals the smallest is positive, the order being total (Ordered field). Order is preserved by adding a constant and by adding inequalities states its moves in their STRICT forms only; the non-strict forms used below follow by adjoining the equality case, in which the two sides coincide, the order being total (Ordered field).

[L4]

Neighbourhoods: $N_{δ}^{*} (c) = {y : 0 < ∣ y - c ∣ < δ}$ , and a smaller radius gives a smaller punctured neighbourhood (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

Proof

technique · direct

1.1L1L3L4choose

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

2.1step 1.1L2L3L4

Let $x \in A$ with $0 < ∣ x - c ∣ < δ$ . Then $0 < ∣ x - c ∣ < δ_{1}$ gives $L - ε < f (x)$ , and $0 < ∣ x - c ∣ < δ_{2}$ gives $h (x) < L + ε$ , while $0 < ∣ x - c ∣ < η$ gives $f (x) \leq g (x) \leq h (x)$ .

3.1step 2.1L2L3

Chaining those four inequalities, $L - ε < f (x) \leq g (x) \leq h (x) < L + ε$ , hence $L - ε < g (x) < L + ε$ , that is $- ε < g (x) - L < ε$ , that is $∣ g (x) - L ∣ < ε$ .

4.1step 3.1L1∎

So for every real $ε > 0$ a real $δ > 0$ has been produced with $∣ g (x) - L ∣ < ε$ for every $x \in A$ satisfying $0 < ∣ x - c ∣ < δ$ : the limit of $g$ at $c$ exists and equals $L$ .

Remarks

Where the three hypotheses are spent. The inequality $f \leq g$ is used only for the lower estimate and $g \leq h$ only for the upper one; the equality of the two outer limits is what makes the two estimates close on the same number $L$ . Drop it and the argument gives only $\lim f \leq lim inf$ -style information, which this page does not develop.
The order hypothesis is local. It is imposed only on $A \cap N_{η}^{*} (c)$ , so the theorem is insensitive to the behaviour of the three functions far from $c$ , and to their values at $c$ ; that is 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 in action.
Typical use. To prove that a bounded oscillating factor is killed by a factor tending to $0$ : if $∣ u (x) ∣ \leq B$ near $c$ then $- B ∣ x - c ∣ \leq (x - c) u (x) \leq B ∣ x - c ∣$ near $c$ , and both outer functions tend to $0$ . That is exactly how $x ψ (1 / x) \to 0$ is proved on the companion page.
The sequential analogue 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 $g$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If f≤g≤hf≤g≤h near cc and ff and hh have the same limit at cc, then so does gg

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

If $f \leq g \leq h$ near $c$ and $f$ and $h$ have the same limit at $c$ , then so does $g$