In calculus, the squeeze theorem, (also known as the pinching theorem or sandwich theorem) is a theorem regarding the limit of a function. This theorem argues that if two functions approach the same limit at a point, and a third function "lies" between those functions; then, the third function also approaches that limit at that point.

If the functions f, g, and h are defined in an interval I containing a except possibly at a itself, and f(x) ≤ g(x) ≤ h(x) for every number x in I for which x ≠ a, and

then .

Example

x² sin(1/x) is "squeezed" by two functions

Consider g(x) = x² sin 1/x.

Trying to calculate the limit of g as x → 0 is difficult by conventional means; substitution will fail since we have a 1/x in the function. Trying to use L'Hôpital's rule fails too; it does not remove the 1/x term. So we turn to using this result.

Let f(x) = -x² and h(x) = x². It can be shown that f(x) and h(x) constitute lower and upper bounds (respectively) to g(x) and thus satisfy f(x) ≤ g(x) ≤ h(x).

We trivially have (because f and h are polynomials)

Given these two conditions, the squeeze theorem states that

Proof of the squeeze theorem

It is given that

so by the definition of the limit of a function at a point, for any ε > 0 there is a δ₁ > 0 such that

if 0 < |x - a| < δ₁ then |f(x) - L| < ε

if 0 < |x - a| < δ₁ then -ε < f(x) - L < ε

if 0 < |x - a| < δ₁ then L - ε < f(x) < L + ε

and a δ₂ > 0 such that for any ε > 0 there is a δ₁ > 0 such that

if 0 < |x - a| < δ₂ then |h(x) - L| < ε and

if 0 < |x - a| < δ₂ then L - ε < h(x) < L + ε.

Then let δ equal the less of δ₁ and δ₂ (δ = min(δ₁, δ₂) ). From the previous statements it follows that

if 0 < |x - a| < δ then L - ε < f(x) and

if 0 < |x - a| < δ then h(x) < L + ε.

It is given that f(x) ≤ g(x) ≤ h(x), so

if 0 < |x - a| < δ then L - ε < f(x) ≤ g(x) ≤ h(x) < L + ε.

if 0 < |x - a| < δ then L - ε < g(x) < L + ε.

if 0 < |x - a| < δ then -ε < g(x) - L < ε.

if 0 < |x - a| < δ then |g(x) - L| < ε.

This fits the definition of a limit for the function g as x approaches a, so .

The sandwich theorem has no relation to the ham sandwich theorem.