Squeeze Theorem and Special Sequence Limits

The squeeze theorem (sandwich theorem) says that if a_n ≤ b_n ≤ c_n for all large n and a_n and c_n both tend to L, then b_n also tends to L. In brief: since −1/n ≤ sin n / n ≤ 1/n and both bounds tend to 0, sin n / n → 0. The same idea works for functions: from cos x ≤ sin x/x ≤ 1 near 0 one gets lim sin x / x = 1.

The squeeze theorem handles limits of expressions that are too complicated for algebra but can be bounded by simpler ones. Together with a short list of special limits (such as ⁿ√n → 1, qⁿ → 0 and (1 + 1/n)ⁿ → e), it solves most limit problems for sequences. This guide states the theorem, proves it, shows techniques for finding bounds, lists the special limits with justifications and gives a table of numerical values.

What is the squeeze theorem?

If a_n ≤ b_n ≤ c_n for all n beyond some index, and lim a_n = lim c_n = L, then lim b_n = L. In particular, the theorem proves the existence of the limit of b_n.

Proof idea: given ε > 0, for large n both a_n and c_n lie in (L − ε, L + ε). Since b_n is between them, L − ε < a_n ≤ b_n ≤ c_n < L + ε, so |b_n − L| &lt. A special case: if |b_n| ≤ c_n and c_n → 0, then b_n → 0.

How do you find the bounds?

• Bounded functions: |sin n| ≤ 1, |cos n| ≤ 1, so |sin n/n| ≤ 1/n.
• Largest and smallest term: a sum of k terms lies between k times the smallest and k times the largest term.
• Roots of sums: ⁿ√(aⁿ + bⁿ) lies between max(a, b) and 2^1/n max(a, b).
• Integer parts: x − 1 < ⌊x⌋ ≤ x.
• Inequalities: n! ≤ nⁿ, 2ⁿ ≥ n, and Bernoulli’s inequality (1 + x)ⁿ ≥ 1 + nx for x ≥ −1.

What are the special limits?

• qⁿ → 0 for |q| < 1. Write |q| = 1/(1 + h) with h > 0; by Bernoulli (1 + h)ⁿ ≥ 1 + nh, so |q|ⁿ ≤ 1/(1 + nh) → 0.
• ⁿ√a → 1 for a > 0. For a > 1, set ⁿ√a = 1 + h_n; then a = (1 + h_n)ⁿ ≥ 1 + nh_n, so 0 < h_n ≤ (a − 1)/n → 0.
• ⁿ√n → 1. Set ⁿ√n = 1 + h_n, so n = (1 + h_n)ⁿ ≥ 1 + n(n − 1)h_n²/2 (binomial theorem), which gives h_n² ≤ 2/n, so h_n → 0.
• n^kqⁿ → 0 for |q| < 1 and any k (exponential beats polynomial).
• aⁿ/n! → 0 for any a (factorial beats exponential).
• (1 + 1/n)ⁿ → e, and (1 + x/n)ⁿ → e^x.
• ln n/n → 0 and n/2ⁿ → 0.

The hierarchy of growth for large n is: ln n ≪ n^p ≪ aⁿ ≪ n! ≪ nⁿ (for a > 1, p > 0), where “≪” means that the ratio of the left to the right tends to 0.

Worked examples

Example 1: a bounded oscillation

(−1)ⁿ/n: −1/n ≤ (−1)ⁿ/n ≤ 1/n, and both bounds tend to 0. So the limit is 0.

Example 2: a sum of terms

1/√(n² + 1) + 1/√(n² + 2) + … + 1/√(n² + n). There are n terms. The smallest is 1/√(n² + n) and the largest is 1/√(n² + 1). So n/√(n² + n) ≤ sum ≤ n/√(n² + 1). Both bounds tend to 1 (divide by n: 1/√(1 + 1/n) and 1/√(1 + 1/n²)). So the limit is 1.

Example 3: root of a sum

ⁿ√(2ⁿ + 3ⁿ): 3 = ⁿ√(3ⁿ) ≤ ⁿ√(2ⁿ + 3ⁿ) ≤ ⁿ√(2 · 3ⁿ) = 3 · 2^1/n → 3. So the limit is 3. More generally, ⁿ√(a₁ⁿ + … + a_kⁿ) → max(a_i).

Example 4: factorial over power

0 ≤ n!/nⁿ = (1/n)(2/n)⋯(n/n) ≤ (1/n) · 1 ⋯ 1 = 1/n → 0. So n!/nⁿ → 0.

Example 5: with the integer part

⌊nx⌋/n for fixed x: nx − 1 < ⌊nx⌋ ≤ nx, so x − 1/n < ⌊nx⌋/n ≤ x. Both bounds tend to x, so the limit is x.

Example 6: a product

(1 − 1/2²)(1 − 1/3²)⋯(1 − 1/n²) = (n + 1)/(2n) → 1/2 (the product telescopes), consistent with the bounds 0 < product ≤ 1 but needing the exact form.

Example 7: the function version

For 0 < x < π/2: cos x < sin x/x < 1 (compare areas of a triangle, a circular sector and a larger triangle). As x → 0, cos x → 1, so sin x/x → 1. And x sin(1/x) → 0 as x → 0, since |x sin(1/x)| ≤ |x|.

Numerical values of special sequences

Where does the theorem come from?

The idea of squeezing a quantity between two others was used by Archimedes to compute π: the perimeters of inscribed and circumscribed polygons bound the circumference of the circle, and as the number of sides grows, both bounds approach the same value. The theorem in its modern form is credited to Carl Friedrich Gauss and appears in rigorous treatments of analysis from the nineteenth century. In several countries it is called the sandwich theorem or the theorem of two policemen.

Where is it used?

The squeeze theorem is used to prove standard limits (sin x/x, the limits of roots), to justify the derivative of trigonometric functions and to show the convergence of sequences defined by inequalities. In probability, it is used for inequalities such as Chebyshev’s, and in numerical analysis for error bounds: if the error is at most 1/n², then it tends to 0.

Which mistakes are common?

Bounds with different limits

Both bounds must tend to the same L. If −1 ≤ (−1)ⁿ ≤ 1, the bounds do not meet, so the theorem gives no limit.

Bounds that do not hold for all large n

The inequalities must hold for all n beyond some index. Check small cases separately.

Using bounds that are too crude

If the upper bound tends to a larger value than the lower bound, refine the bounds. For example, 1/n ≤ sum ≤ n · (1/n) does not give a limit.

Forgetting the absolute value

To show b_n → 0, bound |b_n| by a sequence that tends to 0, as for sin n/n.