Sandwich Theorem and Its Applications
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Direct answer
Trapping a stubborn function between two tame ones that share the same limit forces the stubborn one to that limit: if g(x) ≤ f(x) ≤ h(x) near a point a (except possibly at a itself) and both g and h tend to L, then f tends to L. The theorem powers the most quoted limit in school mathematics — sin x/x → 1 as x → 0, proved by squeezing the sector between two triangles; it handles oscillation-driven pathologies like x sin(1/x) → 0 (bounded by ±|x|); and it resolves greatest-integer expressions such as x[1/x] → 1 as x → 0+, where the inequality [1/x] ≤ 1/x < [1/x] + 1 does the trapping. In sequences, it gives n^(1/n) → 1 and underlies every "bounded times vanishing" argument a JEE paper deploys.
What you must remember
- The statement: g(x) ≤ f(x) ≤ h(x) on a punctured neighbourhood of a, with lim g = lim h = L, implies lim f = L; the value of f at a (or its absence) is irrelevant.
- The sin x/x squeeze: for 0 < x < π/2, sin x < x < tan x (triangle-sector-triangle areas), rearranging to cos x < sin x/x < 1; both ends → 1.
- Bounded times vanishing: |x sin(1/x)| ≤ |x| → 0; more generally (bounded) × (→ 0) → 0 — the workhorse for oscillatory limits.
- Greatest-integer trapping: [1/x] ≤ 1/x < [1/x] + 1 multiplied by x > 0 gives 1 - x < x[1/x] ≤ 1, so x[1/x] → 1 as x → 0+.
- Sequence version: if an ≤ bn ≤ cn eventually and the outer sequences share limit L, then bn → L; applied to 1 ≤ n^(1/n) ≤ ... via the binomial expansion it yields n^(1/n) → 1.
- Fractional-part bound: 0 ≤ x - [x] < 1 always; combined with squeezing it settles one-sided limits of {x} at integers (0 from the right, 1 from the left — so no limit exists).
- Variants that look like squeezes but are not: the mean value theorems bound by derivatives, not by neighbouring functions — do not label LMVT work as sandwich.
One squeeze from start to finish
Prove lim x→0+ x[1/x] = 1. By the defining inequality of the floor, [1/x] ≤ 1/x < [1/x] + 1. Multiply through by x, which is positive near 0+: x[1/x] ≤ 1 < x[1/x] + x, which rearranges to 1 - x < x[1/x] ≤ 1. The lower bound 1 - x tends to 1 and the upper bound is the constant 1, so the trapped function must tend to 1. Notice what happened: the floor's ragged jumps (at every reciprocal integer) never mattered because the squeeze only needs the inequality band, not smoothness — the function oscillates inside a shrinking envelope. The same architecture proves sin x/x: on a unit circle, the triangle of area (1/2)sin x sits inside the sector (1/2)x which sits inside the big triangle (1/2)tan x, so sin x < x < tan x; dividing by sin x gives cos x < x/sin x < 1/cos x, and inverting flips the inequalities to cos x < sin x/x < 1. Both ends go to 1, and the limit — used daily without proof — is established.
How the exam frames it
JEE Main asks for the theorem's statement as a match-or-true/false item and evaluates the standard squeezes as numericals: lim x→0 x sin(1/x) (answer 0), lim x→0 sin 3x/5x (answer 3/5, by the same squeeze rescaled), lim n→∞ (1/2)^n... zero by bounded-vanishing logic. Advanced builds multi-layer versions: lim x→0 x^2 sin(π/x), limits combining floor with trigonometric envelopes, and the classic n-term squeeze lim n→∞ (1! + 2! + ... + n!)^(1/n)/n = 1 for those who extend reading. The dependable traps: claiming lim x→0 sin(1/x) exists by "squeezing" between -1 and 1 — the bounds must converge to a common L, and constants ±1 do not; forgetting the punctured neighbourhood (the trap function may misbehave at the point itself); and inverting inequalities when taking reciprocals (dividing sin x < x < tan x needs care about direction after inversion). The theorem belongs to the limits, continuity and differentiability unit — foundational for both papers.
Frequently asked questions
What is the sandwich (squeeze) theorem?
If g(x) ≤ f(x) ≤ h(x) near a and both g and h converge to the same L, then f converges to L as well — f is forced to the shared limit regardless of its own complexity.
How does the theorem prove sin x/x → 1?
Geometrically, sin x < x < tan x on (0, π/2) rearranges to cos x < sin x/x < 1, and since cos x → 1, the quotient is squeezed to 1.
Why does lim x→0 x sin(1/x) equal 0?
Because |x sin(1/x)| ≤ |x|: an oscillating but bounded factor multiplied by a vanishing one is squeezed to 0.
Does sin(1/x) have a limit as x → 0?
No — it oscillates between -1 and 1, and the sandwich theorem does not apply because the bounds -1 and 1 do not share a common limit.
What does x[1/x] approach as x → 0+?
Exactly 1, since the floor inequality gives 1 - x < x[1/x] ≤ 1 and the envelope collapses to 1.