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.