Squeeze theorem

Illustration of the squeeze theorem

When a sequence lies between two other converging sequences with the same limit, it also converges to this limit.

In calculus, the squeeze theorem (also known as the sandwich theorem, among other namesTemplate:Efn) is a theorem regarding the limit of a function that is bounded between two other functions.

The squeeze theorem is used in calculus and mathematical analysis, typically to confirm the limit of a function via comparison with two other functions whose limits are known. It was first used geometrically by the mathematicians Archimedes and Eudoxus in an effort to compute [[pi|Template:Pi]], and was formulated in modern terms by Carl Friedrich Gauss.

Statement

The squeeze theorem is formally stated as follows. Template:Math theorem

• The functions Template:Mvar and Template:Mvar are said to be lower and upper bounds (respectively) of Template:Mvar.
• Here, Template:Mvar is not required to lie in the interior of Template:Mvar. Indeed, if Template:Mvar is an endpoint of Template:Mvar, then the above limits are left- or right-hand limits.
• A similar statement holds for infinite intervals: for example, if Template:Math, then the conclusion holds, taking the limits as Template:Math.

This theorem is also valid for sequences. Let Template:Math be two sequences converging to Template:Mvar, and Template:Math a sequence. If $\mathrm{\forall }n\ge N,N\in \mathbb{N}$ we have Template:Math, then Template:Math also converges to Template:Mvar.

Proof

According to the above hypotheses we have, taking the limit inferior and superior: $$L=\underset{x\to a}{\lim }g(x)\le \underset{x\to a}{\mathrm{lim\; inf}}f(x)\le \underset{x\to a}{\mathrm{lim\; sup}}f(x)\le \underset{x\to a}{\lim }h(x)=L,$$ so all the inequalities are indeed equalities, and the thesis immediately follows.

A direct proof, using the Template:Math-definition of limit, would be to prove that for all real Template:Math there exists a real Template:Math such that for all Template:Mvar with $|x-a|<\delta ,$ we have $|f(x)-L|<\epsilon .$ Symbolically,

$$\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0:\mathrm{\forall }x,(|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon ).$$

As

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

means that Template:NumBlk

and $$\underset{x\to a}{\lim }h(x)=L$$

means that Template:NumBlk

then we have

$$g(x)\le f(x)\le h(x)$$ $$g(x)-L\le f(x)-L\le h(x)-L$$

We can choose $\delta :=\min \{{\delta }_1,{\delta }_2\}$. Then, if $|x-a|<\delta$, combining (Template:EquationNote) and (Template:EquationNote), we have

$$-\epsilon <g(x)-L\le f(x)-L\le h(x)-L<\epsilon ,$$ $$-\epsilon <f(x)-L<\epsilon ,$$

which completes the proof. Q.E.D

The proof for sequences is very similar, using the $\epsilon$-definition of the limit of a sequence.

Examples

First example

$x^2\sin \left(\frac{1}{x}\right)$ being squeezed in the limit as Template:Mvar goes to 0

The limit

$$\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{1}{x}\right)$$

cannot be determined through the limit law

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

because

$$\underset{x\to 0}{\lim }\sin \left(\frac{1}{x}\right)$$

does not exist.

However, by the definition of the sine function,

$$-1\le \sin \left(\frac{1}{x}\right)\le 1.$$

It follows that

$$-x^2\le x^2\sin \left(\frac{1}{x}\right)\le x^2$$

Since $\underset{x\to 0}{\lim }-x^2=\underset{x\to 0}{\lim }{x}^2=0$, by the squeeze theorem, $\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{1}{x}\right)$ must also be 0.

Second example

Comparing areas:
$\begin{array}{cccccc}& A(\mathrm{△}ADB)& \le & A(\text{sector }ADB)& \le & A(\mathrm{△}ADF)\\ \Rightarrow & \frac{1}{2}\cdot \sin x\cdot 1& \le & \frac{x}{2\pi }\cdot \pi & \le & \frac{1}{2}\cdot \tan x\cdot 1\\ \Rightarrow & \sin x& \le & x& \le & \frac{\sin x}{\cos x}\\ \Rightarrow & \frac{\cos x}{\sin x}& \le & \frac{1}{x}& \le & \frac{1}{\sin x}\\ \Rightarrow & \cos x& \le & \frac{\sin x}{x}& \le & 1\end{array}$

Probably the best-known examples of finding a limit by squeezing are the proofs of the equalities $$\begin{array}{cc}& \underset{x\to 0}{\lim }\frac{\sin x}{x}=1,\\ & \underset{x\to 0}{\lim }\frac{1-\cos x}{x}=0.\end{array}$$

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

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

for Template:Mvar close enough to 0. The correctness of which for positive Template:Mvar can be seen by simple geometric reasoning (see drawing) that can be extended to negative Template:Mvar as well. The second limit follows from the squeeze theorem and the fact that

$$0\le \frac{1-\cos x}{x}\le x$$ for Template:Mvar close enough to 0. This can be derived by replacing Template:Math in the earlier fact by $\sqrt{1-{\cos }^{2}x}$ and squaring the resulting inequality.

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

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

since the radius is Template:Math and the arc on the unit circle has length Template:Math. Similarly, the area of the larger of the two shaded sectors is

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

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 Template:Math, and the height is 1. The area of the triangle is therefore

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

From the inequalities

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

we deduce that

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

provided Template:Math, and the inequalities are reversed if Template:Math. Since the first and third expressions approach Template:Math as Template:Math, and the middle expression approaches $\frac{d}{d\theta }\tan \theta ,$ 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.

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

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

$$\begin{array}{cccccc}& 0& \le & \frac{x^2}{x^2+y^2}& \le & 1\\ -|y|\le y\le |y|⟹& -|y|& \le & \frac{x^{2}y}{x^2+y^2}& \le & |y|\\ \genfrac{}{}{0ex}{}{\underset{(x,y)\to (0,0)}{\lim }-|y|=0}{\underset{(x,y)\to (0,0)}{\lim }|y|=0}⟹& 0& \le & \underset{(x,y)\to (0,0)}{\lim }\frac{x^{2}y}{x^2+y^2}& \le & 0\end{array}$$

therefore, by the squeeze theorem,

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

External links

• Template:MathWorld
• Squeeze Theorem by Bruce Atwood (Beloit College) after work by, Selwyn Hollis (Armstrong Atlantic State University), the Wolfram Demonstrations Project.
• Squeeze Theorem on ProofWiki.