The Order Limit Theorem and Squeeze

Suppose you know that one sequence lies between two others. Suppose both outer sequences settle down to the same number. Must the trapped sequence settle down to that number too? This question sounds almost too obvious to need a proof. Yet making it rigorous forces us to confront exactly what a limit is. It rewards us with one of the most useful tools in all of analysis: the Squeeze Theorem. The deeper issue underneath this question is whether the order structure of the real numbers survives the passage to a limit. This structure is the relation ≤\le. Limits are defined by an ε\varepsilon-bookkeeping process. It is not a priori clear that an inequality true for every term of a sequence must remain true. This holds even non-strictly, for the limiting value. The Order Limit Theorem answers this precisely. It is the engine that makes the Squeeze Theorem work. These results matter because direct computation of a limit is often hopeless. Comparison to simpler, already-understood sequences, by contrast, is easy. One bounds an unwieldy expression between two tractable ones. Then one lets the bounds converge together. This technique is used everywhere. Examples include estimating sin⁡(x)/x\sin(x)/x near 00, and proving n1/n→1n^{1/n} \to 1. It also helps in bounding error terms in numerical analysis. It controls remainder terms in Taylor's theorem too. This article gives the exact statement and complete proof of the Order Limit Theorem. It derives the Squeeze Theorem as a clean corollary. It also works through the subtleties. Especially important is why strict inequalities can degrade to non-strict ones in the limit. Equally important is why every hypothesis is truly necessary. This includes convergence of the bounding sequences, and eventual truth of the inequality.

Key takeaways
  • The Order Limit Theorem says that inequalities between sequences pass to their limits, but strict inequalities can only guarantee non-strict inequalities in the limit.
  • It requires both sequences to actually converge — inequalities alone say nothing about limits that don't exist.
  • The inequality only needs to hold eventually (for all nn past some point), not for every single term.
  • The Squeeze Theorem is the corollary where a sequence trapped between two sequences with a common limit is forced to share that limit.
  • Classic uses: proving n1/n→1n^{1/n}\to 1, sin⁡nn→0\frac{\sin n}{n}\to 0, and bounding sequences that are too messy to compute directly.
  • Key idea: trap the unknown sequence between two known ones, and let the trap close.

Where this lives today

The Squeeze Theorem generalizes immediately to functions and continuous limits (x→ax \to a, x→∞x \to \infty), where it is a workhorse for computing limits like lim⁡x→0x2sin⁡(1/x)=0\lim_{x\to 0} x^2 \sin(1/x) = 0 or lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 via geometric bounding arguments. In real analysis, it underlies convergence tests for series (comparison test), estimates for integrals (bounding an integrand above and below), and error analysis in numerical methods, where truncation or discretization error is squeezed between explicit bounds that both vanish as a step size shrinks. In probability and statistics, squeeze-type arguments bound tail probabilities or estimators between simpler quantities whose limiting behavior is known (a technique central to concentration inequalities). In computer science, the same idea appears in algorithm analysis: bounding a recurrence or runtime function between two simpler asymptotic bounds (as in Θ\Theta-notation, which is literally a two-sided squeeze). The Order Limit Theorem itself is the quiet foundation beneath the Monotone Convergence Theorem and the study of Cauchy Sequences and Completeness, both of which rely on comparing limiting behavior via inequalities that must be handled with exactly this level of care.

Ask the tutor: Can you show me how the Squeeze Theorem is used to prove that lim⁡n→∞n!nn=0\lim_{n\to\infty} \frac{n!}{n^n} = 0?

DefinitionConvergent sequence, limit, and order comparison

Let (an)n=1∞(a_n)_{n=1}^{\infty} be a sequence of real numbers and L∈RL \in \mathbb{R}. We say (an)(a_n) converges to LL, written

lim⁡n→∞an=L,\lim_{n\to\infty} a_n = L,

if for every ε>0\varepsilon > 0 there exists N∈NN \in \mathbb{N} such that

n≥N  ⟹  ∣an−L∣<ε.n \ge N \implies |a_n - L| < \varepsilon.

For two sequences (an)(a_n) and (bn)(b_n), we say an≤bna_n \le b_n eventually if there exists M∈NM \in \mathbb{N} such that an≤bna_n \le b_n for all n≥Mn \ge M. (If the inequality holds for every n∈Nn \in \mathbb{N} we simply say it holds "for all nn"; this is the special case M=1M = 1.)

Intuition

Picture a sequence (an)(a_n) as a moving point on the number line, and a limit LL as a fixed target that the point approaches, getting arbitrarily close and staying close forever. If an≤bna_n \le b_n for every (or every sufficiently large) index nn, then the moving point ana_n never gets ahead of bnb_n. As n→∞n \to \infty, both points converge — ana_n toward L1L_1 and bnb_n toward L2L_2 — and the "never gets ahead" relationship is inherited by the limit points: L1≤L2L_1 \le L_2. But notice the subtlety: even if an<bna_n < b_n strictly for every nn, the gap bn−anb_n - a_n can shrink to zero, so all we can conclude is L1≤L2L_1 \le L_2, not L1<L2L_1 < L_2. A classic instance: an=0a_n = 0 and bn=1/nb_n = 1/n satisfy an<bna_n < b_n for all nn, yet both limits equal 00. The Squeeze Theorem is the natural next question: if a sequence cnc_n is sandwiched, an≤cn≤bna_n \le c_n \le b_n, and the two outer sequences converge to the same limit LL, then cnc_n has no room to do anything except also converge to LL. The "sandwich" metaphor is exact — as the bounds slam together at LL, the middle sequence is crushed into converging there too.

Figure
Two convergent sequences with an≤bna_n \le b_n, and their limits satisfying L1≤L2L_1 \le L_2
n value L₁ L₂ aₙ bₙ aₙ ≤ bₙ for all n ⇒ L₁ ≤ L₂
TheoremOrder Limit Theorem

Let (an)(a_n) and (bn)(b_n) be sequences of real numbers such that

lim⁡n→∞an=L1andlim⁡n→∞bn=L2\lim_{n\to\infty} a_n = L_1 \qquad \text{and} \qquad \lim_{n\to\infty} b_n = L_2

both exist. Then: 1. If an≥0a_n \ge 0 eventually, then L1≥0L_1 \ge 0. 2. If an≤bna_n \le b_n eventually, then L1≤L2L_1 \le L_2. 3. If there exists c∈Rc \in \mathbb{R} with an≤ca_n \le c eventually, then L1≤cL_1 \le c. Similarly, if an≥ca_n \ge c eventually, then L1≥cL_1 \ge c.

Proof

We use direct ε\varepsilon-arguments from the definition of limit (proof by contradiction for the main claim).

Part 1. Suppose an≥0a_n \ge 0 for all n≥Mn \ge M, for some M∈NM \in \mathbb{N}. We claim L1≥0L_1 \ge 0. Suppose toward contradiction that L1<0L_1 < 0. Let ε=−L1>0\varepsilon = -L_1 > 0. Since an→L1a_n \to L_1, there exists N∈NN \in \mathbb{N} such that

n≥N  ⟹  ∣an−L1∣<ε=−L1.n \ge N \implies |a_n - L_1| < \varepsilon = -L_1.

Take any n≥max⁡(M,N)n \ge \max(M, N). The inequality ∣an−L1∣<−L1|a_n - L_1| < -L_1 unpacks to

L1−(−L1)<an<L1+(−L1)=0,L_1 - (-L_1) < a_n < L_1 + (-L_1) = 0,

i.e. an<0a_n < 0. But n≥Mn \ge M forces an≥0a_n \ge 0 by hypothesis — a contradiction. Hence no such nn can exist consistently, so our assumption L1<0L_1 < 0 is false, and L1≥0L_1 \ge 0.

Part 2. Suppose an≤bna_n \le b_n for all n≥Mn \ge M. Define cn=bn−anc_n = b_n - a_n. By the Algebraic Limit Theorem (limit of a difference), (cn)(c_n) converges, and

lim⁡n→∞cn=L2−L1.\lim_{n\to\infty} c_n = L_2 - L_1.

By hypothesis, cn=bn−an≥0c_n = b_n - a_n \ge 0 for all n≥Mn \ge M, so cn≥0c_n \ge 0 eventually. Applying Part 1 to the sequence (cn)(c_n) gives L2−L1≥0L_2 - L_1 \ge 0, that is,

L1≤L2.L_1 \le L_2.

Part 3. The statement "an≤ca_n \le c eventually" is the special case of Part 2 obtained by taking bn=cb_n = c for all nn (the constant sequence), which trivially converges to L2=cL_2 = c by definition of limit (for any ε>0\varepsilon>0, ∣bn−c∣=0<ε|b_n - c| = 0 < \varepsilon for all nn). Part 2 then gives L1≤cL_1 \le c. The second claim (an≥ca_n \ge c eventually implies L1≥cL_1 \ge c) follows identically by taking bn=anb_n = a_n, an′=ca_n' = c constant, and applying Part 2 in the form c≤an  ⟹  c≤L1c \le a_n \implies c \le L_1. ■\blacksquare

∎
CorollarySqueeze Theorem

Let (an)(a_n), (bn)(b_n), (cn)(c_n) be sequences of real numbers such that: 1. an≤cn≤bna_n \le c_n \le b_n for all nn sufficiently large (eventually), and 2. lim⁡n→∞an=lim⁡n→∞bn=L\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = L for some common L∈RL \in \mathbb{R}. Then (cn)(c_n) converges and lim⁡n→∞cn=L\lim_{n\to\infty} c_n = L.

Proof

We argue directly from the ε\varepsilon–NN definition of limit; this proof is independent of the Order Limit Theorem (we do not need to know in advance that (cn)(c_n) converges — we prove it from scratch). Let ε>0\varepsilon > 0 be given. Since an→La_n \to L, there exists N1∈NN_1 \in \mathbb{N} such that

n≥N1  ⟹  ∣an−L∣<ε,i.e.L−ε<an<L+ε.n \ge N_1 \implies |a_n - L| < \varepsilon, \quad \text{i.e.} \quad L - \varepsilon < a_n < L + \varepsilon.

Since bn→Lb_n \to L, there exists N2∈NN_2 \in \mathbb{N} such that

n≥N2  ⟹  ∣bn−L∣<ε,i.e.L−ε<bn<L+ε.n \ge N_2 \implies |b_n - L| < \varepsilon, \quad \text{i.e.} \quad L - \varepsilon < b_n < L + \varepsilon.

By hypothesis (1), there exists M∈NM \in \mathbb{N} such that an≤cn≤bna_n \le c_n \le b_n for all n≥Mn \ge M. Let N=max⁡(N1,N2,M)N = \max(N_1, N_2, M). For all n≥Nn \ge N, all three conditions hold simultaneously, so we may chain the inequalities:

L−ε<an≤cn≤bn<L+ε.L - \varepsilon < a_n \le c_n \le b_n < L + \varepsilon.

Extracting the outer terms, L−ε<cn<L+εL - \varepsilon < c_n < L + \varepsilon, i.e. ∣cn−L∣<ε|c_n - L| < \varepsilon. Since ε>0\varepsilon > 0 was arbitrary, and we produced NN (dependent on ε\varepsilon) such that n≥N  ⟹  ∣cn−L∣<εn \ge N \implies |c_n - L| < \varepsilon, we conclude by definition that

lim⁡n→∞cn=L.■\lim_{n\to\infty} c_n = L. \qquad \blacksquare
∎

Worked examples

Example 1: Proving lim⁡n→∞cos⁡(n3)+7n2n+sin⁡(n2)=72\displaystyle\lim_{n\to\infty} \frac{\cos(n^3) + 7n}{2n + \sin(n^2)} = \frac{7}{2} is false to compute directly, but bounding works. This sequence is too wild to attack with algebraic limit rules alone because cos⁡(n3)\cos(n^3) and sin⁡(n2)\sin(n^2) oscillate with no pattern amenable to direct simplification. We squeeze instead. For all n≥1n \ge 1: since −1≤cos⁡(n3)≤1-1 \le \cos(n^3) \le 1 and −1≤sin⁡(n2)≤1-1 \le \sin(n^2) \le 1, we have for the numerator

7n−1≤cos⁡(n3)+7n≤7n+1,7n - 1 \le \cos(n^3) + 7n \le 7n + 1,

and for the denominator, since 2n≥2>02n \ge 2 > 0 for n≥1n \ge 1, and −1≤sin⁡(n2)≤1-1 \le \sin(n^2) \le 1,

2n−1≤2n+sin⁡(n2)≤2n+1.2n - 1 \le 2n + \sin(n^2) \le 2n + 1.

Both 2n−1>02n - 1 > 0 for n≥1n \ge 1, so the denominator is always positive and we may divide inequalities safely. Using

7n−12n+1≤cos⁡(n3)+7n2n+sin⁡(n2)≤7n+12n−1\frac{7n-1}{2n+1} \le \frac{\cos(n^3)+7n}{2n+\sin(n^2)} \le \frac{7n+1}{2n-1}

(this chain requires justification: since the denominator 2n+sin⁡(n2)∈[2n−1,2n+1]2n+\sin(n^2) \in [2n-1, 2n+1] is positive, and the numerator is in [7n−1,7n+1][7n-1, 7n+1], the fraction is minimized by taking the smallest numerator over the largest denominator, and maximized by the largest numerator over the smallest denominator — both endpoints are legitimate since all quantities are positive for n≥1n \ge 1). Now compute the two bounding limits using the Algebraic Limit Theorem:

lim⁡n→∞7n−12n+1=lim⁡n→∞7−1/n2+1/n=7−02+0=72,\lim_{n\to\infty} \frac{7n-1}{2n+1} = \lim_{n\to\infty} \frac{7 - 1/n}{2 + 1/n} = \frac{7 - 0}{2+0} = \frac{7}{2},
lim⁡n→∞7n+12n−1=lim⁡n→∞7+1/n2−1/n=7+02−0=72.\lim_{n\to\infty} \frac{7n+1}{2n-1} = \lim_{n\to\infty} \frac{7 + 1/n}{2 - 1/n} = \frac{7+0}{2-0} = \frac{7}{2}.

Both outer sequences converge to 7/27/2. By the Squeeze Theorem,

lim⁡n→∞cos⁡(n3)+7n2n+sin⁡(n2)=72.\lim_{n\to\infty} \frac{\cos(n^3)+7n}{2n+\sin(n^2)} = \frac{7}{2}.

Example 2: Proving lim⁡n→∞n1/n=1\displaystyle\lim_{n\to\infty} n^{1/n} = 1. This is a genuinely nontrivial limit — no algebraic limit law applies directly since both the base and exponent vary with nn. We squeeze using the binomial theorem. For n≥1n \ge 1, write n1/n=1+dnn^{1/n} = 1 + d_n where dn≥0d_n \ge 0 (this is valid since n1/n≥1n^{1/n} \ge 1 for all n≥1n \ge 1, because n≥1n \ge 1). We seek an upper bound on dnd_n. For n≥2n \ge 2, raise both sides to the nnth power:

n=(1+dn)n.n = (1+d_n)^n.

By the binomial theorem, since all terms are non-negative (as dn≥0d_n \ge 0), we may discard all but one term to get a valid lower bound on the right side. Using the k=2k=2 term:

(1+dn)n=∑k=0n(nk)dnk≥(n2)dn2=n(n−1)2dn2.(1+d_n)^n = \sum_{k=0}^n \binom{n}{k} d_n^k \ge \binom{n}{2} d_n^2 = \frac{n(n-1)}{2} d_n^2.

Therefore

n≥n(n−1)2dn2  ⟹  dn2≤2nn(n−1)=2n−1(n≥2),n \ge \frac{n(n-1)}{2} d_n^2 \implies d_n^2 \le \frac{2n}{n(n-1)} = \frac{2}{n-1} \quad (n \ge 2),

so

0≤dn≤2n−1for all n≥2.0 \le d_n \le \sqrt{\frac{2}{n-1}} \qquad \text{for all } n \ge 2.

Now apply the Squeeze Theorem to (dn)(d_n): the lower bound sequence is constantly 00, and

lim⁡n→∞2n−1=2⋅lim⁡n→∞1n−1=2⋅0=0,\lim_{n\to\infty} \sqrt{\frac{2}{n-1}} = \sqrt{2}\cdot \lim_{n\to\infty}\frac{1}{\sqrt{n-1}} = \sqrt2 \cdot 0 = 0,

using that 1/n−1→01/\sqrt{n-1} \to 0 (a standard consequence of the Archimedean property together with the algebraic limit theorem for the composition with ⋅\sqrt{\cdot}, or directly from the definition). Hence dn→0d_n \to 0 by the Squeeze Theorem. Finally, n1/n=1+dnn^{1/n} = 1 + d_n, so by the Algebraic Limit Theorem (limit of a sum),

lim⁡n→∞n1/n=1+lim⁡n→∞dn=1+0=1.\lim_{n\to\infty} n^{1/n} = 1 + \lim_{n\to\infty} d_n = 1 + 0 = 1.
CounterexampleWhy the sequences must actually converge

The Order Limit Theorem requires that lim⁡an\lim a_n and lim⁡bn\lim b_n both exist. Without this, the conclusion can fail entirely — not just numerically, but in the sense that "the limit" doesn't even exist to compare. Take an=(−1)na_n = (-1)^n and bn=(−1)n+1nb_n = (-1)^n + \frac{1}{n}. Then for every n≥1n \ge 1,

an<bna_n < b_n

since 1/n>01/n > 0. This inequality is not just eventual but holds for every index. Yet neither (an)(a_n) nor (bn)(b_n) converges: ana_n oscillates permanently between −1-1 and 11 (for even nn, an=1a_n = 1; for odd nn, an=−1a_n = -1), and bnb_n does the same up to a vanishing correction. There is no L1L_1 or L2L_2 for the theorem's conclusion L1≤L2L_1 \le L_2 to even refer to. This shows the hypothesis "(an)(a_n) and (bn)(b_n) converge" cannot be dropped or replaced by something weaker — the theorem is not merely about the existence of an order relation between terms, but about transporting that relation through an actual limiting process, which requires the limits to exist in the first place. Note also that even a strict, uniform inequality (an<bna_n < b_n for all nn) gives no information whatsoever once convergence fails.

Scope, edge cases & common traps

SituationWhat is trueWhat is not true
an≤bna_n \le b_n eventually, both convergentL1≤L2L_1 \le L_2L1<L2L_1 < L_2 (even if an<bna_n < b_n strictly)
an<bna_n < b_n for all nnstill only L1≤L2L_1 \le L_2strict inequality of limits
an≤cn≤bna_n \le c_n \le b_n, an,bn→La_n, b_n \to Lcn→Lc_n \to L (Squeeze)conclusion if an,bna_n, b_n have different limits
an≤bna_n \le b_n for infinitely many nn (not eventually)nothing guaranteedorder theorem does not apply
(an),(bn)(a_n), (b_n) don't convergeno conclusion about limitsany comparison of "limits" is meaningless

The mathematician behind it

Origins · 1 minAugustin-Louis Cauchy and how this was discovered
AC
Augustin-Louis Cauchy
1789–1857 · Foundations of rigorous analysis in early 19th-century France

Cauchy was one of the principal architects of the modern, rigorous theory of limits, sequences, and series, formalized in his 1821 lecture notes Cours d'Analyse at the École Polytechnique. Before Cauchy (and later Weierstrass, who gave the definitive ε\varepsilon–δ\delta/ε\varepsilon–NN formalism), mathematicians manipulated infinite processes with intuitive but logically shaky arguments about "infinitely small quantities." Cauchy's insistence on precise inequalities to define convergence — the direct ancestor of the definitions used above — made results like the Order Limit Theorem and Squeeze Theorem provable rather than merely plausible.

History
- 1821 — Cauchy's Cours d'Analyse gives an early rigorous treatment of limits and convergence criteria for sequences and series.
- 1837 — Bolzano's posthumously influential work on real analysis anticipates key completeness ideas underlying limit comparisons.
- 1850s–1860s — Weierstrass develops and popularizes the ε\varepsilon–δ\delta (and ε\varepsilon–NN) formalism, giving the modern precise language used to state and prove the Order Limit Theorem.
- 1872 — Dedekind and Cantor construct rigorous foundations for the real numbers, securing the completeness properties that make comparisons of limits fully consistent.
- Late 19th–20th century — the Squeeze Theorem becomes a standard staple of calculus and analysis curricula worldwide, taught as an elementary but indispensable comparison technique.
Why can't strict inequalities between terms give strict inequalities between limits?

Because convergence only controls the eventual behavior of a sequence, not the exact value of the gap between terms at any finite stage. If an<bna_n < b_n for every nn, the difference bn−anb_n - a_n is some positive number at each step, but that positive number is allowed to shrink toward 00 as n→∞n \to \infty. The classic example is an=0a_n = 0 and bn=1/nb_n = 1/n: strictly 0<1/n0 < 1/n for every nn, yet both converge to the same value 00. In the limit, the strict gap is squeezed away entirely, so the Order Limit Theorem can only certify L1≤L2L_1 \le L_2, and this bound is tight — it cannot be strengthened without extra information.

Do the outer sequences in the Squeeze Theorem need to converge to the same limit?

Yes, this is essential, not optional. If an→L1a_n \to L_1 and bn→L2b_n \to L_2 with L1≠L2L_1 \ne L_2, then a sequence cnc_n satisfying an≤cn≤bna_n \le c_n \le b_n is only forced to satisfy L1≤lim inf⁡cn≤lim sup⁡cn≤L2L_1 \le \liminf c_n \le \limsup c_n \le L_2 (using more advanced tools like limsup/liminf), and cnc_n need not converge at all — it might oscillate anywhere in [L1,L2][L_1, L_2]. The entire force of the Squeeze Theorem comes from the outer bounds collapsing onto a single point, leaving cnc_n no room to do anything else.

Does the inequality in these theorems need to hold for every single nn, or just eventually?

Just eventually — that is, for all nn greater than or equal to some fixed threshold MM. This is because convergence itself is entirely about the tail behavior of a sequence; finitely many exceptional terms at the start have no effect on whether or where a sequence converges. Both the Order Limit Theorem and Squeeze Theorem are stated and proved using this weaker, more flexible "eventual" hypothesis, which makes them considerably easier to apply, since one rarely wants to verify an inequality for small, awkward initial indices.

Can the Squeeze Theorem be used to prove a sequence diverges?

Not directly in its stated form, since its conclusion is always convergence to a specific common limit LL. However, a closely related comparison idea — sometimes called the divergence comparison test — can show divergence: if an≤cna_n \le c_n eventually and an→∞a_n \to \infty, then cn→∞c_n \to \infty as well. This is a natural companion result proved by a nearly identical direct argument from the definition of divergence to infinity, but it is logically distinct from the Squeeze Theorem itself, which strictly requires both bounding sequences to converge to the same finite limit.

How does this relate to limits of functions, as opposed to sequences?

The sequential Squeeze Theorem is the direct ancestor of the function version: if g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) for all xx near aa (except possibly at aa itself), and lim⁡x→ag(x)=lim⁡x→ah(x)=L\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L, then lim⁡x→af(x)=L\lim_{x \to a} f(x) = L. In fact, using the sequential characterization of function limits, the function version can be derived from the sequence version: take any sequence xn→ax_n \to a with xn≠ax_n \ne a, apply the sequential Squeeze Theorem to g(xn)≤f(xn)≤h(xn)g(x_n) \le f(x_n) \le h(x_n), and conclude f(xn)→Lf(x_n) \to L for every such sequence, which is precisely the definition of lim⁡x→af(x)=L\lim_{x\to a} f(x) = L.

Is the Order Limit Theorem the same as saying limits preserve order isomorphisms?

No — it is a much more modest and specific statement. It says only that the non-strict order relation ≤\le passes from eventual term-wise comparison to the corresponding limits, when those limits exist. It says nothing about strict order, about sequences that don't converge, or about more exotic order-theoretic structures. It is best understood as a basic compatibility fact between the algebraic/topological notion of limit and the order structure of R\mathbb{R}, not a deep structural isomorphism theorem.

Before this

Where this leads