Squeeze theorem

File:(x^2)sin(x^(-1)).png

Illustration of the squeeze theorem

File:Sandwich lemma.svg

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.<ref>Template:Cite book</ref> 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 <math>\forall n\geq N, N\in\N</math> 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: <math display="block">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,</math> 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 <math>|x - a| < \delta,</math> we have <math>|f(x) - L| < \varepsilon.</math> Symbolically,

<math display="block"> \forall \varepsilon > 0, \exists \delta > 0 : \forall x, (|x - a | < \delta \ \Rightarrow |f(x) - L |< \varepsilon).</math>

As

<math display="block">\lim_{x \to a} g(x) = L </math>

means that Template:NumBlk

and <math display="block">\lim_{x \to a} h(x) = L </math>

means that Template:NumBlk

then we have

<math display="block">g(x) \leq f(x) \leq h(x) </math> <math display="block">g(x) - L\leq f(x) - L\leq h(x) - L</math>

We can choose <math>\delta:=\min\left\{\delta_1,\delta_2\right\}</math>. Then, if <math>|x - a| < \delta</math>, combining (Template:EquationNote) and (Template:EquationNote), we have

<math display="block"> - \varepsilon < g(x) - L\leq f(x) - L\leq h(x) - L\ < \varepsilon, </math> <math display="block"> - \varepsilon < f(x) - L < \varepsilon ,</math>

which completes the proof. Q.E.D

The proof for sequences is very similar, using the <math>\varepsilon</math>-definition of the limit of a sequence.

Examples

First example

File:Inst satsen.svg

<math>x^2 \sin\left(\tfrac{1}{x}\right)</math> being squeezed in the limit as Template:Mvar goes to 0

The limit

<math display="block">\lim_{x \to 0}x^2 \sin\left( \tfrac{1}{x} \right)</math>

cannot be determined through the limit law

<math display="block">\lim_{x \to a}(f(x) \cdot g(x)) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x),</math>

because

<math display="block">\lim_{x\to 0}\sin\left( \tfrac{1}{x} \right)</math>

does not exist.

However, by the definition of the sine function,

<math display="block">-1 \le \sin\left( \tfrac{1}{x} \right) \le 1. </math>

It follows that

<math display="block">-x^2 \le x^2 \sin\left( \tfrac{1}{x} \right) \le x^2 </math>

Since <math>\lim_{x\to 0}-x^2 = \lim_{x\to 0}x^2 = 0</math>, by the squeeze theorem, <math>\lim_{x\to 0} x^2 \sin\left(\tfrac{1}{x}\right)</math> must also be 0.

Second example

File:Limit sin x x.svg

Comparing areas:
<math>\begin{array}{cccccc} & A(\triangle ADB) & \leq & A(\text{sector } ADB) & \leq & A(\triangle ADF) \\[4pt] \Rightarrow & \frac{1}{2} \cdot \sin x \cdot 1 & \leq & \frac{x}{2\pi} \cdot \pi & \leq & \frac{1}{2} \cdot \tan x \cdot 1 \\[4pt] \Rightarrow & \sin x & \leq & x & \leq & \frac{\sin x}{\cos x} \\[4pt] \Rightarrow & \frac{\cos x}{\sin x} & \leq & \frac{1}{x} & \leq & \frac{1}{\sin x} \\[4pt] \Rightarrow & \cos x & \leq & \frac{\sin x}{x} & \leq & 1 \end{array}</math>

Probably the best-known examples of finding a limit by squeezing are the proofs of the equalities <math display="block"> \begin{align} & \lim_{x\to 0} \frac{\sin x}{x} =1, \\[10pt] & \lim_{x\to 0} \frac{1 - \cos x}{x} = 0. \end{align} </math>

The first limit follows by means of the squeeze theorem from the fact that<ref>Selim G. Krejn, V.N. Uschakowa: Vorstufe zur höheren Mathematik. Springer, 2013, Template:ISBN, pp. 80-81 (German). See also Sal Khan: Proof: limit of (sin x)/x at x=0 (video, Khan Academy)</ref>

<math display="block"> \cos x \leq \frac{\sin x}{x} \leq 1 </math>

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

<math display="block"> 0 \leq \frac{1 - \cos x}{x} \leq x </math> for Template:Mvar close enough to 0. This can be derived by replacing Template:Math in the earlier fact by <math display="inline"> \sqrt{1-\cos^2 x}</math> 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 <math display="block"> \frac{d}{d\theta} \tan\theta = \sec^2\theta </math> by squeezing, as follows.

File:Tangent.squeeze.svg

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

<math display="block"> \frac{\sec^2\theta\,\Delta\theta}{2}, </math>

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

<math display="block"> \frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2}. </math>

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

<math display="block"> \frac{\tan(\theta + \Delta\theta) - \tan\theta}{2}. </math>

From the inequalities

<math display="block"> \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} </math>

we deduce that

<math display="block"> \sec^2\theta \le \frac{\tan(\theta + \Delta\theta) - \tan\theta}{\Delta\theta} \le \sec^2(\theta + \Delta\theta),</math>

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 <math>\tfrac{d}{d\theta} \tan\theta,</math> 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.<ref>Template:Cite book</ref>

<math display="block">\lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2}</math>

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

<math display="block">\begin{array}{rccccc}

 & 0 & \leq & \displaystyle \frac{x^2}{x^2+y^2} & \leq & 1 \\[4pt]
 -|y| \leq y \leq |y| \implies & -|y| & \leq & \displaystyle \frac{x^2 y}{x^2+y^2} & \leq & |y| \\[4pt]
 {
   {\displaystyle \lim_{(x,y) \to (0, 0)} -|y| = 0} \atop 
   {\displaystyle \lim_{(x,y) \to (0, 0)} \ \ \ |y| = 0}
 } \implies & 0 & \leq & \displaystyle \lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2} & \leq & 0
 \end{array}</math> 

therefore, by the squeeze theorem,

<math display="block">\lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2} = 0.</math>

References

References

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External links

• {{#invoke:Template wrapper|{{#if:|list|wrap}}|_template=cite web

|_exclude=urlname, _debug, id |url = https://mathworld.wolfram.com/{{#if:SqueezeTheorem%7CSqueezeTheorem.html}} |title = Squeeze Theorem |author = Weisstein, Eric W. |website = MathWorld |access-date = |ref = Template:SfnRef }}

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