Squeeze theorem

From Exampleproblems

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

The squeeze theorem is a technical result which is very important in proofs in calculus and mathematical analysis. It is typically used to confirm the limit of a function via comparison with two other functions whose limits are known or easily computed. It was first used geometrically by the mathematicians Archimedes and Eudoxus in an effort to compute π. It was formulated in modern terms by Gauss.

In Italian, the squeeze theorem is also known as the two carabineri theorem. The idea is that two carabinieri are holding a prisoner in the middle.

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

1 Statement
2 Examples and applications
3 Proof

Statement

The squeeze theorem is formally stated as follows.

Let I be an interval containing the point a. Let f, g, and h be functions defined on I, except possibly at a itself. Suppose that for every x in I not equal to a, we have
$g(x) \leq f(x) \leq h(x)$ <p> and also suppose that <p> $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L.$ <p> Then $\lim_{x \to a} f(x) = L$ . </blockquote>
The functions g(x) and h(x) are said to be lower and upper bounds (respectively) of f(x).
Note that a is not required to lie in the interior of I. Indeed, if a is an endpoint of I, then the above limits are left- or right-hand limits.
A similar statement holds for infinite intervals: for example, if I = (0, ∞), then the conclusion holds, taking the limits as x → ∞.

Examples and applications

The following examples illustrate how the squeeze theorem is applied in practice.

Example 1

Image:Squeeze theorem example.png
x² sin(1/x) is "squeezed" by two functions

Consider f(x) = x² sin(1/x), which is defined on any interval containing x = 0, but is not defined at x = 0 itself.
Computing the limit of f(x) as x → 0 is difficult by conventional means. Direct substitution fails because the function is not defined at x = 0, (let alone continuous). We cannot use L'Hopital's rule either, because the sin(1/x) factor always oscillates and fails to settle on a limit. However, the squeeze theorem works with suitably chosen functions.
Since the sine function is always bounded in absolute value by 1, it follows that f(x) is bounded in absolute value by the function x². In other words, letting g(x) = −x² and h(x) = x², we have

$g(x) \leq f(x) \leq h(x).$

Since g and h are polynomials, they are continuous, and so

$\lim_{x \to 0} g(x) = \lim_{x \to 0} h(x) = 0$

By the squeeze theorem, we conclude

$\lim_{x \to 0} f(x) = 0.$

Example 2

The above example is a particular application of a general situation which occurs frequently. Suppose we wish to show that

$\lim_{x \to a} f(x) = L.$

Then it is sufficient to find a function h(x), defined on some interval I containing a, except possibly at a itself, such that

$\lim_{x \to a} h(x) = 0$

and such that for all x in I not equal to a,

$|f(x) - L| \leq h(x).$

In words, this means that the error between f(x) and L can be made arbitrarily small, provided we take x close enough to a.
To see that these conditions are sufficient, note that since the absolute value function is always non-negative, we may take g(x) = 0 for all x and apply the squeeze theorem: as $x \rightarrow a, \, |f(x) - L| \rightarrow 0$ , so that $f(x) - L \rightarrow 0$ , which implies that

$f(x) = (f(x) - L) + L \rightarrow 0 + L = L.$

Example 3

Using elementary geometry, one can show that for 0 < x < π/2, the following inequality holds:

$\cos x < \frac{\sin x}{x} < 1.$

Since

$\lim_{x \to 0} \cos x = 1$

the squeeze theorem tells us that

$\lim_{x \to 0} \frac{\sin x}{x} = 1.$

This limit is instrumental in computing the derivative of the sine function.

Proof

The main idea behind this proof is to consider the relative differences between the functions f, g, and h. This has the effect of making the lower bound identically 0, and all the functions non-negative. This greatly simplifies the details of the proof. The general case then follows algebraically.
To begin the proof, assume all the hypotheses and notation as given in the statement of the theorem above. We first prove the special case where g(x) = 0 for all x and L = 0. In this case,

$\lim_{x \to a} h(x) = 0.$

Let ε > 0 be any fixed positive number. By the definition of the limit of a function, there is a δ > 0 such that

$\mbox{if }0 < |x - a| < \delta, \mbox{ then }|h(x)| < \varepsilon.$

For any x in I not equal to a,

$0 = g(x) \leq f(x) \leq h(x)$

so that

$|f(x)| \leq |h(x)|.$

We conclude that

$\mbox{if }0 < |x - a| < \delta, \mbox{ then }|f(x)| \leq |h(x)| < \varepsilon.$

This proves that

$\lim_{x \to a} f(x) = 0 = L.$

This completes the proof for the special case. Now, we prove the general theorem by letting g and L be arbitrary. For any x in I not equal to a, we have

$g(x) \leq f(x) \leq h(x).$

Subtracting g(x) from each expression,

$0 \leq f(x) - g(x) \leq h(x) - g(x).$

As $x \rightarrow a, \, h(x) \rightarrow L$ and $g(x) \rightarrow L$ , so that

$h(x) - g(x) \rightarrow L - L = 0.$

The special case now shows that $f(x) - g(x) \rightarrow 0.$ We conclude that

$f(x) = (f(x) - g(x)) + g(x) \rightarrow 0 + L = L.$
This completes the proof. Q.E.D.