The squeeze theorem

Statement

Let \( (a_n)_{n \in \mathbb{N}} \), \( (b_n)_{n \in \mathbb{N}} \) and \( (c_n)_{n \in \mathbb{N}} \) be sequences of real numbers, and let \( L \in \mathbb{R} \). Suppose that (i) there exists \( N_0 \in \mathbb{N} \) such that \( a_n \le b_n \le c_n \) for all \( n \ge N_0 \); (ii) \( a_n \to L \) as \( n \to \infty \); and (iii) \( c_n \to L \) as \( n \to \infty \), with the same limit \( L \) in both cases. Then \( (b_n) \) converges and \( \lim_{n \to \infty} b_n = L \).

Why it matters

Most sequences met in practice have no closed form for their terms — sums, oscillating products, terms defined by inequalities. The squeeze theorem converts the problem of computing a limit into the far easier problem of finding two-sided estimates: if you can trap an intractable sequence between two tractable ones sharing a limit, you are done. It is the standard engine behind limits such as \( \frac{\sin n}{n} \to 0 \), \( n^{1/n} \to 1 \), and Riemann-sum estimates, and its estimation philosophy is the germ of the comparison test for series and, much later, of dominated convergence.

Conceptually it is the first theorem in analysis that exploits the order structure of \( \mathbb{R} \) rather than its algebra: it shows that the limit operation respects the order relation in a strong, two-sided way. Notably, its proof uses no completeness — it is valid verbatim in \( \mathbb{Q} \) — which places it at a different foundational level from, say, the monotone convergence theorem.

Hypotheses

Proof

Result

Corollaries & converses

• Absolute-value form. If \( |b_n| \le c_n \) eventually and \( c_n \to 0 \), then \( b_n \to 0 \). (Apply the theorem with \( a_n = -c_n \), noting \( -c_n \to 0 \) by the algebra of limits.) This is the form used in nine out of ten applications.
• Metric form. If \( |b_n - L| \le c_n \) eventually and \( c_n \to 0 \), then \( b_n \to L \); this version generalises verbatim to any metric space via \( d(b_n, L) \le c_n \).
• Order-limit corollary. Taking \( b_n = c_n \): if \( a_n \le c_n \) eventually and both converge, then \( \lim a_n \le \lim c_n \) — weak inequalities survive passage to the limit (strict ones do not: \( \frac{1}{n} \gt 0 \) but the limit is \( 0 \)).
• Converse (outer from inner) is false. If \( a_n \le b_n \le c_n \) and \( b_n \to L \), the outer sequences need not converge: \( a_n = -2 + (-1)^n \le 0 = b_n \le 2 + (-1)^n = c_n \), yet neither \( (a_n) \) nor \( (c_n) \) converges.
• Trivial converse holds vacuously. Every convergent sequence is squeezed (take \( a_n = c_n = b_n \)), so the theorem exactly characterises nothing beyond convergence itself; its value is practical, not classificatory.

Fails without

Common errors

• The one-sided squeeze. Bounding only from above — "\( b_n \le \frac{1}{n} \to 0 \), so \( b_n \to 0 \)" — is invalid: \( b_n = -n \) satisfies the bound. You need both fences, or the absolute-value form \( |b_n| \le c_n \).
• Forgetting to check the bounds converge. Trapping \( b_n \) between \( (-1)^n - \frac{1}{n} \) and \( (-1)^n + \frac{1}{n} \) proves nothing, however tight the trap: the fences themselves diverge.
• Strictness leakage. From \( a_n \lt b_n \lt c_n \) concluding \( L \lt \lim b_n \lt L \) — strict inequalities do not survive limits; the conclusion is \( \lim b_n = L \), full stop.
• Squeezing with unequal limits and "averaging". If \( a_n \to 1 \) and \( c_n \to 3 \), the middle sequence does not converge to \( 2 \); it need not converge at all.
• Circularity. Using a bound whose proof already requires the limit being computed — e.g. bounding \( n^{1/n} \) by an expression whose limit is evaluated using \( n^{1/n} \to 1 \). Each fence's limit must be established independently.
• Misreading "eventually". Students discard the theorem because the sandwich fails at \( n = 1, 2, 3 \). Finitely many violations are irrelevant: only behaviour beyond some \( N_0 \) matters.

Discussion

The squeeze principle is ancient in spirit: Archimedes' method of exhaustion traps the area of the disc between inscribed and circumscribed polygons whose areas he could compute, obtaining \( 3\tfrac{10}{71} \lt \pi \lt 3\tfrac{1}{7} \). This is exactly a squeeze avant la lettre — two computable sequences converging on an incomputable target. The modern \( \varepsilon \)–\( N \) formulation belongs to the nineteenth-century rigourisation of analysis (Bolzano, Cauchy, and in final polished form Weierstrass), where it appears as one of the first theorems provable directly from the definition of limit, requiring no completeness axiom.

Structurally, the theorem is the sharpest elementary expression of the compatibility between the order and the topology of \( \mathbb{R} \). Its two-sided nature is essential and reflects the fact that intervals \( (L - \varepsilon, L + \varepsilon) \) form a neighbourhood base: convergence in \( \mathbb{R} \) is eventual confinement to shrinking order-intervals, and the sandwich hypothesis hands the middle sequence exactly that confinement. Its descendants are everywhere: the comparison test for series (squeeze the partial sums), the sandwich theorem for function limits \( x \to a \) (via the sequential characterisation of limits), and error estimates of the form \( |b_n - L| \le C r^n \) throughout numerical analysis, where the squeeze converts a computable error bound into a convergence proof — this is the "change" thread's basic template: control a process by controlling its distance to the target.

Two refinements deserve note. First, the theorem is order-theoretic, not metric: in any linearly ordered set equipped with the order topology, if \( a_n \le b_n \le c_n \) eventually and \( a_n, c_n \to L \), then \( b_n \to L \), because every order-neighbourhood of \( L \) contains an interval \( (p, q) \) with \( p \lt L \lt q \), and eventually \( p \lt a_n \le b_n \le c_n \lt q \). No distances, no completeness, no field structure. Second, the clean route through upper and lower limits: for real sequences, \( a_n \le b_n \le c_n \) eventually gives \( \liminf a_n \le \liminf b_n \) and \( \limsup b_n \le \limsup c_n \) (monotonicity of \( \liminf \) and \( \limsup \) under eventual domination), whence \( L \le \liminf b_n \le \limsup b_n \le L \), forcing \( \liminf b_n = \limsup b_n = L \), which is equivalent to convergence to \( L \). This second proof needs completeness (to define \( \limsup \) via suprema) but generalises smoothly — it is the pattern behind Fatou's lemma sandwiching in measure theory, where dominated convergence is precisely an infinite-dimensional squeeze with the dominating function as the fence. In spaces without order ("space" thread), the surviving form is the metric estimate \( d(b_n, L) \le c_n \to 0 \): the squeeze retreats to the real-valued distance sequence, which is where the order lives.

Common misconceptions. The theorem does not require the sandwich from \( n = 1 \) (eventual suffices); it does not require the fences to be monotone or to approach \( L \) from opposite sides (both may oscillate through \( L \)); it does not use the completeness of \( \mathbb{R} \) (the \( \varepsilon \)–\( N \) proof above is valid over \( \mathbb{Q} \)); and it gives no information whatever when the fences have different limits — not even boundedness-based partial conclusions about the limit's location beyond \( \liminf a_n \le \liminf b_n \) and \( \limsup b_n \le \limsup c_n \).

Worked examples

Example 1. Show that \( \displaystyle \lim_{n \to \infty} \frac{\sin n}{n} = 0 \).

Example 2. Show that \( \displaystyle \lim_{n \to \infty} n^{1/n} = 1 \).

Problems

1. Prove that \( \displaystyle \lim_{n\to\infty} \frac{3 + \cos(n^2)}{n} = 0 \).
   Solution

   Since \( -1 \le \cos(n^2) \le 1 \) (boundedness of cosine), we have \( 2 \le 3 + \cos(n^2) \le 4 \), hence for all \( n \ge 1 \):

   \[ \frac{2}{n} \le \frac{3 + \cos(n^2)}{n} \le \frac{4}{n} . \]

   Both \( \frac{2}{n} \to 0 \) and \( \frac{4}{n} \to 0 \) (constant multiples of \( \frac{1}{n} \to 0 \), by the algebra of limits and the Archimedean property). By the squeeze theorem with \( L = 0 \), the middle sequence converges to \( 0 \). \( \blacksquare \)

2. Prove that \( \displaystyle \lim_{n\to\infty} \left( 2^n + 5^n \right)^{1/n} = 5 \). You may use \( x^{1/n} \to 1 \) for fixed \( x \gt 0 \).
   Solution

   For every \( n \ge 1 \): from \( 5^n \le 2^n + 5^n \le 2 \cdot 5^n \), taking \( n \)-th roots (the map \( t \mapsto t^{1/n} \) is increasing on \( [0,\infty) \)) gives

   \[ 5 \le \left( 2^n + 5^n \right)^{1/n} \le 2^{1/n} \cdot 5 . \]

   The left fence is constantly \( 5 \). The right fence: \( 2^{1/n} \to 1 \) by the cited fact, so \( 2^{1/n} \cdot 5 \to 5 \) by the product rule for limits. By the squeeze theorem, \( (2^n + 5^n)^{1/n} \to 5 \). \( \blacksquare \)

   Remark: the same trick shows \( (x_1^n + \dots + x_k^n)^{1/n} \to \max_i x_i \) for nonnegative reals — the "\( \ell^p \to \ell^\infty \)" phenomenon in miniature.

3. Prove that \( \displaystyle \lim_{n\to\infty} \sum_{k=1}^{n} \frac{n}{n^2 + k} = 1 \).
   Solution

   Each summand is squeezed by replacing \( k \) with its extreme values \( 1 \le k \le n \): for every \( k \in \{1, \dots, n\} \),

   \[ \frac{n}{n^2 + n} \le \frac{n}{n^2 + k} \le \frac{n}{n^2 + 1} . \]

   Summing \( n \) copies of each fence:

   \[ \frac{n^2}{n^2 + n} \le \sum_{k=1}^{n} \frac{n}{n^2 + k} \le \frac{n^2}{n^2 + 1} . \]

   For the left fence, \( \frac{n^2}{n^2+n} = \frac{1}{1 + 1/n} \to \frac{1}{1+0} = 1 \) by the algebra of limits (quotient rule, denominator limit nonzero). For the right fence, \( \frac{n^2}{n^2+1} = \frac{1}{1 + 1/n^2} \to 1 \) likewise. By the squeeze theorem the sum converges to \( 1 \). \( \blacksquare \)

4. Let \( (b_n) \) be a real sequence, \( L \in \mathbb{R} \), and suppose there exists \( r \in [0, 1) \) with \( |b_{n+1} - L| \le r\,|b_n - L| \) for all \( n \in \mathbb{N} \). Prove \( b_n \to L \). You may use \( r^n \to 0 \) for \( 0 \le r \lt 1 \).
   Solution

   Claim: \( |b_n - L| \le r^{n-1} |b_1 - L| \) for all \( n \ge 1 \). Proof by induction: the base case \( n = 1 \) reads \( |b_1 - L| \le |b_1 - L| \). If the claim holds for \( n \), then

   \[ |b_{n+1} - L| \le r\,|b_n - L| \le r \cdot r^{n-1} |b_1 - L| = r^{n} |b_1 - L| , \]

   using the hypothesis and the induction hypothesis (multiplication by \( r \ge 0 \) preserves \( \le \)). This closes the induction.

   Now squeeze:

   \[ 0 \le |b_n - L| \le r^{n-1} |b_1 - L| . \]

   The left fence is constantly \( 0 \to 0 \); the right fence tends to \( 0 \) since \( r^{n-1} \to 0 \) (cited) and constants factor out by the algebra of limits. By the squeeze theorem, \( |b_n - L| \to 0 \).

   Finally, \( |b_n - L| \to 0 \iff b_n \to L \): both statements read "\( \forall \varepsilon \gt 0 \; \exists N \; \forall n \ge N : |b_n - L| \lt \varepsilon \)", since \( \bigl| |b_n - L| - 0 \bigr| = |b_n - L| \). Hence \( b_n \to L \). \( \blacksquare \)

   Remark: this is the convergence engine inside the contraction mapping theorem.

5. (Structural.) For bounded real sequences define \( \limsup_{n} x_n := \lim_{n\to\infty} \sup_{k \ge n} x_k \) and \( \liminf_n x_n := \lim_{n\to\infty} \inf_{k \ge n} x_k \). (a) Prove the monotonicity lemma: if \( x_n \le y_n \) for all \( n \ge N_0 \), then \( \limsup_n x_n \le \limsup_n y_n \) and \( \liminf_n x_n \le \liminf_n y_n \). (b) Using (a) and the fact that \( x_n \to L \iff \liminf x_n = \limsup x_n = L \), give a second proof of the squeeze theorem for bounded sequences. (c) Explain where this proof uses completeness of \( \mathbb{R} \), and why the \( \varepsilon \)–\( N \) proof does not.
   Solution

   (a) Fix \( n \ge N_0 \). For every \( k \ge n \) we have \( x_k \le y_k \le \sup_{j \ge n} y_j \), so \( \sup_{j \ge n} y_j \) is an upper bound for \( \{x_k : k \ge n\} \); by definition of supremum as least upper bound, \( \sup_{k \ge n} x_k \le \sup_{k \ge n} y_k \). The sequences \( s_n^x := \sup_{k\ge n} x_k \) and \( s_n^y := \sup_{k \ge n} y_k \) are decreasing and bounded, hence convergent (monotone convergence theorem), and \( s_n^x \le s_n^y \) for \( n \ge N_0 \) passes to the limit by the order-limit corollary: \( \limsup x_n \le \limsup y_n \). The \( \liminf \) statement is dual (or apply the \( \limsup \) result to \( -y_n \le -x_n \), using \( \liminf x_n = -\limsup(-x_n) \)).

   (b) Assume \( a_n \le b_n \le c_n \) for \( n \ge N_0 \) with \( a_n \to L \), \( c_n \to L \), all sequences bounded. By (a) applied twice:

   \[ L = \liminf_n a_n \le \liminf_n b_n \le \limsup_n b_n \le \limsup_n c_n = L , \]

   where the outer equalities use \( a_n \to L \iff \liminf a_n = \limsup a_n = L \) (cited equivalence), and the middle inequality \( \liminf \le \limsup \) holds because \( \inf_{k \ge n} b_k \le \sup_{k \ge n} b_k \) for each \( n \) and weak inequalities pass to limits. All four quantities are therefore equal to \( L \), so \( \liminf b_n = \limsup b_n = L \), and by the cited equivalence \( b_n \to L \). \( \blacksquare \)

   (c) The \( \limsup \) proof needs \( \sup_{k \ge n} x_k \) to exist, which is the completeness (least-upper-bound) axiom of \( \mathbb{R} \) — in \( \mathbb{Q} \), a bounded set need not have a supremum, and \( \limsup \) may be undefined. The \( \varepsilon \)–\( N \) proof manipulates only the order axioms and the definition of limit, never invoking suprema or completeness; it is therefore valid in \( \mathbb{Q} \) and in any ordered field. The squeeze theorem is genuinely more elementary than the \( \limsup \) machinery used in this alternative proof.