# Squeeze theorem

"Sandwich theorem" redirects here. For the result in measure theory, see Ham sandwich theorem.

In calculus, the squeeze theorem (known also as the pinching theorem, the sandwich theorem, the sandwich rule and sometimes the squeeze lemma) is a theorem regarding the limit of a function.

The squeeze theorem is a technical result that 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, and was formulated in modern terms by Gauss.

In Italy, China, Chile, Russia, Poland, Hungary and France, the squeeze theorem is also known as the two carabinieri theorem, two militsioner theorem, sandwich theorem, two gendarmes theorem, "Double sided theorem" or two policemen and a drunk theorem. The story is that if two policemen are escorting a drunk prisoner between them, and both officers go to a cell, then (regardless of the path taken, and the fact that the prisoner may be wobbling about between the policemen) the prisoner must also end up in the cell.

## Statement

The squeeze theorem is formally stated as follows.

Let I be an interval having the point a as a limit point. 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:

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

and also suppose that:

$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L. \,$

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

• The functions g and h are said to be lower and upper bounds (respectively) of f.
• Here 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 → ∞.

### Proof

From the above hypotheses we have, taking the limit inferior and superior:

$L=\lim_{x \to a} g(x)\leq \liminf_{x\to a}f(x)\leq\limsup_{x\to a}f(x)\leq \lim_{x \to a}h(x)=L,$

so all the inequalities are indeed equalities, and the thesis immediately follows.

Another proof, using the (ε, δ) definition of limit, would be to prove that for all real ε > 0 there exists a real δ > 0 such that for all x with 0 < |xa | < δ, we have −ε < f(x) − L < ε. Symbolically,

$\forall \varepsilon > 0\ \exists \ \delta > 0 : \forall x\ (0 < |x - a | < \delta \ \Rightarrow \ - \varepsilon < f(x) - L < \varepsilon).$.

As

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

means that

$\forall \varepsilon > 0\ \exists \ \delta_1 > 0 : \forall x\ (0 < |x - a| < \delta_1 \ \Rightarrow \ - \varepsilon < g(x) - L < \varepsilon).\qquad (1)$

and

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

means that

$\forall \varepsilon > 0\ \exists \ \delta_2 > 0 : \forall x\ (0 < |x - a | < \delta_2\ \Rightarrow \ - \varepsilon < h(x) - L < \varepsilon),\qquad (2)$

then we have

$g(x) \leq f(x) \leq h(x) \,$
$g(x) - L\leq f(x) - L\leq h(x) - L\,$

We can choose $\delta$ such that $\delta<\delta_1$ and $\delta<\delta_2$; e.g., by choosing $\delta:=\frac{1}{2}\min\left\{\delta_1,\delta_2\right\}$. Then, if $|x - a|<\delta$, combining (1) and (2), we have

$- \varepsilon < g(x) - L\leq f(x) - L\leq h(x) - L\ < \varepsilon,$
$- \varepsilon < f(x) - L < \varepsilon$,

which completes the proof. $\blacksquare$

## Examples

### First example

x2 sin(1/x) being squeezed in the limit as x goes to 0

The limit

$\lim_{x \to 0}x^2 \sin(\tfrac{1}{x})$

cannot be ascertained through the limit law

$\lim_{x \to a}(f(x)\cdot g(x)) = \lim_{x \to a}f(x)\cdot \lim_{x \to a}g(x),$

because

$\lim_{x\to 0}\sin(\tfrac{1}{x})$

does not exist.

However, by the definition of the sine function,

$-1 \le \sin(\tfrac{1}{x}) \le 1. \,$

It follows that

$-x^2 \le x^2 \sin(\tfrac{1}{x}) \le x^2 \,$

Since $\lim_{x\to 0}-x^2 = \lim_{x\to 0}x^2 = 0$, by the squeeze theorem, $\lim_{x\to 0} x^2 \sin(\tfrac{1}{x})$ must also be 0.

### Second example

Probably the best-known examples of finding a limit by squeezing are the proofs of the equalities

\begin{align} & \lim_{x\to 0} \frac{\sin x}{x} =1, \\[10pt] & \lim_{x\to 0} \frac{1 - \cos x}{x} = 0. \end{align}

The first follows by means of the squeeze theorem from the fact that

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

for x close enough, but not equal to 0.

These two limits are used in proofs of the fact that the derivative of the sine function is the cosine function. That fact is relied on in other proofs of derivatives of trigonometric functions.

### Third example

It is possible to show that

$\frac{d}{d\theta} \tan\theta = \sec^2\theta$

by squeezing, as follows.

In the illustration at right, the area of the smaller of the two shaded sectors of the circle is

$\frac{\sec^2\theta\,\Delta\theta}{2},$

since the radius is sec θ and the arc on the unit circle has length Δθ. Similarly the area of the larger of the two shaded sectors is

$\frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2}.$

What is squeezed between them is the triangle whose base is the vertical segment whose endpoints are the two dots. The length of the base of the triangle is tan(θ + Δθ) − tan(θ), and the height is 1. The area of the triangle is therefore

$\frac{\tan(\theta + \Delta\theta) - \tan(\theta)}{2}.$

From the inequalities

$\frac{\sec^2\theta\,\Delta\theta}{2} \le \frac{\tan(\theta + \Delta\theta) - \tan(\theta)}{2} \le \frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2}$

we deduce that

$\sec^2\theta \le \frac{\tan(\theta + \Delta\theta) - \tan(\theta)}{\Delta\theta} \le \sec^2(\theta + \Delta\theta),$

provided Δθ > 0, and the inequalities are reversed if Δθ < 0. Since the first and third expressions approach sec2θ as Δθ → 0, and the middle expression approaches (d/) tan θ, the desired result follows.

### Fourth example

The squeeze theorem can still be used in multivariable calculus but the lower (and upper functions) must be below (and above) the target function not just along a path but around the entire neighborhood of the point of interest and it only works if the function really does have a limit there. It can, therefore, be used to prove that a function has a limit at a point, but it can never be used to prove that a function does not have a limit at a point.[1]

$\lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2}$

cannot be found by taking any number of limits along paths that pass through the point, but since

$0 \leq \frac{x^2}{x^2+y^2} \leq 1$
$-\left | y \right \vert \leq y \leq \left | y \right \vert$
$-\left | y \right \vert \leq \frac{x^2 y}{x^2+y^2} \leq \left | y \right \vert$
$\lim_{(x,y) \to (0, 0)} -\left | y \right \vert = 0$
$\lim_{(x,y) \to (0, 0)} \left |y \right \vert = 0$
$0 \leq \lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2} \leq 0$

therefore, by the squeeze theorem,

$\lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2} = 0$

## References

1. ^ Stewart, James (2008). "Chapter 15.2 Limits and Continuity". Multivariable Calculus (6th ed.). pp. 909–910. ISBN 0495011630.