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

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

Statement

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

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

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 $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ 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 $x \mapsto x\,\psi(1/x)$ on the companion page is, and the theorem still delivers its limit at $c$ .

The proof is a direct $\varepsilon$ - $\delta$ argument and uses no choice principle.

Facts & Assumptions

Given: A set $A \subseteq \mathbb{R}$ , a limit point $c$ of $A$ , functions $f, g, h : A \to \mathbb{R}$ , a real $\eta > 0$ with $f(x) \le g(x) \le h(x)$ for every $x \in A$ satisfying $0 < |x - c| < \eta$ , and a real $L$ with $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} h(x) = L$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$ ).

[L1]

The limit condition: for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that every $x \in A$ with $0 < |x - c| < \delta$ satisfies $|f(x) - L| < \varepsilon$ , and likewise for $h$ (The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ 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 $\mathbb{R}$ : the order is transitive, and mixed chains compose, so $u < v \le w$ gives $u < w$ and $u \le v < w$ 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: $N^{*}_{\delta}(c) = \{\, y : 0 < |y - c| < \delta \,\}$ , and a smaller radius gives a smaller punctured neighbourhood (The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

Proof

technique · direct

1.1

Let $\varepsilon > 0$ be an arbitrary real. By [L1] fix reals $\delta_1, \delta_2 > 0$ such that every $x \in A$ with $0 < |x - c| < \delta_1$ satisfies $|f(x) - L| < \varepsilon$ and every $x \in A$ with $0 < |x - c| < \delta_2$ satisfies $|h(x) - L| < \varepsilon$ ; let $\delta$ be the smallest of $\delta_1$ , $\delta_2$ and $\eta$ , so $\delta > 0$ .

L1L3L4choose

2.1

Let $x \in A$ with $0 < |x - c| < \delta$ . Then $0 < |x - c| < \delta_1$ gives $L - \varepsilon < f(x)$ , and $0 < |x - c| < \delta_2$ gives $h(x) < L + \varepsilon$ , while $0 < |x - c| < \eta$ gives $f(x) \le g(x) \le h(x)$ .

step 1.1L2L3L4

3.1

Chaining those four inequalities, $L - \varepsilon < f(x) \le g(x) \le h(x) < L + \varepsilon$ , hence $L - \varepsilon < g(x) < L + \varepsilon$ , that is $-\varepsilon < g(x) - L < \varepsilon$ , that is $|g(x) - L| < \varepsilon$ .

step 2.1L2L3

4.1

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

step 3.1L1∎

Remarks

Where the three hypotheses are spent. The inequality $f \le g$ is used only for the lower estimate and $g \le h$ 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 $\lim f \le \liminf$ -style information, which this page does not develop.
The order hypothesis is local. It is imposed only on $A \cap N^{*}_{\eta}(c)$ , 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 $|u(x)| \le B$ near $c$ then $-B|x - c| \le (x - c)u(x) \le B|x - c|$ near $c$ , and both outer functions tend to $0$ . That is exactly how $x\,\psi(1/x) \to 0$ is proved on the companion page.
The sequential analogue is The squeeze theorem.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 29 results over 12 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)
Squeeze theorem (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≤g≤hf \le g \le hf≤g≤h near ccc and fff and hhh have the same limit at ccc, then so does ggg