TheoremStatement: AI-adaptedProof: Literature-sourcedprecheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)

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The squeeze theorem

Statement

Let $(a_{k})$ , $(x_{k})$ and $(b_{k})$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) with

$a_{k} \leq x_{k} \leq b_{k} eventually,$

and suppose $(a_{k})$ and $(b_{k})$ both converge to the same real $L$ (Limits and Cauchy sequences of reals). Then $(x_{k})$ converges to $L$ .

The middle sequence is not assumed to converge; that is the point of the theorem, and it is why the squeeze is a tool for establishing convergence rather than for computing a limit already known to exist.

Facts & Assumptions

Given: Sequences $(a_{k})$ , $(x_{k})$ , $(b_{k})$ of reals, an index $K_{0} \in N$ with $a_{k} \leq x_{k} \leq b_{k}$ for every $k \geq K_{0}$ , and a real $L$ such that $(a_{k})$ converges to $L$ and $(b_{k})$ converges to $L$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L1]

Convergence, quantified over rational $ε > 0$ (Limits and Cauchy sequences of reals).

[L2]

Absolute value: for $c > 0$ , $∣ u ∣ < c$ if and only if $- c < u < c$ (Basic properties of the absolute value).

[L3]

Order arithmetic in $R$ : adding a constant preserves $<$ and $\leq$ ; $u < v \leq w$ and $u \leq v < w$ both give $u < w$ ; and $\leq$ is transitive (Order is preserved by adding a constant and by adding inequalities, Complete ordered field (least-upper-bound property), Ordered field).

[L4]

The order on $N$ is total and transitive, so three indices $K_{0}, K_{1}, K_{2}$ admit an index $K$ with $K \geq K_{0}$ , $K \geq K_{1}$ and $K \geq K_{2}$ , and then $k \geq K$ gives $k \geq K_{0}$ , $k \geq K_{1}$ , $k \geq K_{2}$ ( $\leq$ is a linear order on $N$ ).

Proof

technique · direct

1.1L1

Let $ε > 0$ be rational. By convergence of $(a_{k})$ there is $K_{1}$ with $∣ a_{k} - L ∣ < ε$ for all $k \geq K_{1}$ , and by convergence of $(b_{k})$ there is $K_{2}$ with $∣ b_{k} - L ∣ < ε$ for all $k \geq K_{2}$ .

2.1step 1.1L2L3

By [L2], $∣ a_{k} - L ∣ < ε$ gives $- ε < a_{k} - L$ , that is $L - ε < a_{k}$ , for all $k \geq K_{1}$ ; and $∣ b_{k} - L ∣ < ε$ gives $b_{k} - L < ε$ , that is $b_{k} < L + ε$ , for all $k \geq K_{2}$ .

3.1step 2.1givenL2L3L4choose

Choose $K$ with $K \geq K_{0}$ , $K \geq K_{1}$ and $K \geq K_{2}$ . For every $k \geq K$ the hypothesis gives $a_{k} \leq x_{k} \leq b_{k}$ , so $L - ε < a_{k} \leq x_{k} \leq b_{k} < L + ε$ , hence $L - ε < x_{k} < L + ε$ , hence $- ε < x_{k} - L < ε$ , hence $∣ x_{k} - L ∣ < ε$ .

4.1step 3.1L1∎

The rational $ε > 0$ was arbitrary, so for every rational $ε > 0$ there is an index $K$ beyond which $∣ x_{k} - L ∣ < ε$ ; that is, $(x_{k})$ converges to $L$ .

Remarks

The proof is direct from the definition and does not route through Limits preserve non-strict inequalities. It could not: that lemma assumes the middle sequence converges, which is what is being proved here.
Both bounding sequences must have the same limit. With $a_{k} = - 1$ , $b_{k} = 1$ and any $(x_{k})$ taking values in $[- 1, 1]$ the hypothesis $a_{k} \leq x_{k} \leq b_{k}$ holds and nothing whatever follows, since a bounded sequence need not converge (FALSE: every bounded sequence converges).
The most common use is with $a_{k} = - c_{k}$ and $b_{k} = c_{k}$ for a null sequence $(c_{k})$ : if $∣ x_{k} ∣ \leq c_{k}$ eventually and $c_{k} \to 0$ , then $x_{k} \to 0$ . That special case also follows from A null sequence times a bounded sequence is null whenever $x_{k}$ is presented as a product of a null sequence and a bounded one, but the squeeze needs no such presentation.

The squeeze theorem

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources