Editing Squeeze theorem

{{Short description|Method for finding limits in calculus}}
{{more citations needed|date=April 2010}}
{{Redirect|Sandwich theorem|the result in measure theory|Ham sandwich theorem|Sandwich theory (physics)|Sandwich theory}}

[[File:(x^2)sin(x^(-1)).png|thumb|300px|Illustration of the squeeze theorem]]
[[File:Sandwich lemma.svg|thumb|300px|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 names{{efn|Also known as the ''pinching theorem'', the ''sandwich rule'', the ''police theorem'', the ''between theorem'' and sometimes the ''squeeze lemma''.}}) is a [[theorem]] regarding the [[limit of a function|limit]] of a [[Function (mathematics)|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 of Cnidus|Eudoxus]] in an effort to compute [[pi|{{pi}}]], and was formulated in modern terms by [[Carl Friedrich Gauss]].

== Statement ==
The squeeze theorem is formally stated as follows.<ref>{{cite book|last1=Sohrab|first1=Houshang H.|title=Basic Real Analysis| date=2003|publisher=[[Birkhäuser]]|isbn=978-1-4939-1840-9|page=104|edition=2nd|url=https://books.google.com/books?id=QnpqBQAAQBAJ&pg=PA104}}</ref>
{{math theorem|
Let {{mvar|I}} be an [[interval (mathematics)|interval]] containing the point {{mvar|a}}.  Let {{mvar|g}}, {{mvar|f}}, and {{mvar|h}} be [[function (mathematics)|functions]] defined on {{mvar|I}}, except possibly at {{mvar|a}} itself. Suppose that for every {{mvar|x}} in {{mvar|I}} not equal to {{mvar|a}}, we have
<math display="block">g(x) \leq f(x) \leq h(x) </math>
and also suppose that
<math display="block">\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L. </math>
Then <math>\lim_{x \to a} f(x) = L.</math>
}}

* The functions {{mvar|g}} and {{mvar|h}} are said to be [[upper bound|lower and upper bounds]] (respectively) of {{mvar|f}}.
* Here, {{mvar|a}} is ''not'' required to lie in the [[interior (topology)|interior]] of {{mvar|I}}. Indeed, if {{mvar|a}} is an endpoint of {{mvar|I}}, then the above limits are left- or right-hand limits.
* A similar statement holds for infinite intervals: for example, if {{math|1=''I'' = (0, &infin;)}}, then the conclusion holds, taking the limits as {{math|''x'' → &infin;}}.
This theorem is also valid for sequences. Let {{math|(''a{{sub|n}}''), (''c{{sub|n}}'')}} be two sequences converging to {{mvar|ℓ}}, and {{math|(''b{{sub|n}}'')}} a sequence. If <math>\forall n\geq N, N\in\N</math> we have {{math|''a{{sub|n}}'' ≤ ''b{{sub|n}}'' ≤  ''c{{sub|n}}''}}, then {{math|(''b{{sub|n}}'')}} also converges to {{mvar|ℓ}}.

===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 {{math|(''&epsilon;'', ''&delta;'')}}-definition of limit, would be to prove that for all real {{math|''&epsilon;'' > 0}} there exists a real {{math|''&delta;'' > 0}} such that for all {{mvar|x}} 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
{{NumBlk||<math display="block"> \forall \varepsilon > 0, \exists \ \delta_1 > 0 : \forall x\ (|x - a| < \delta_1 \ \Rightarrow \ |g(x) - L |< \varepsilon).</math>|{{EquationRef|1}}}}

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

means that
{{NumBlk||<math display="block"> \forall \varepsilon > 0, \exists \ \delta_2 > 0 : \forall x\ (|x - a | < \delta_2\ \Rightarrow \ |h(x) - L |< \varepsilon), </math>|{{EquationRef|2}}}}

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 ({{EquationNote|1}}) and ({{EquationNote|2}}), 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|thumb|right|250px|<math>x^2 \sin\left(\tfrac{1}{x}\right)</math> being squeezed in the limit as {{mvar|x}} 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|thumb|upright=1.5|Comparing areas:<br/>
<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, {{ISBN|9783322986283}}, pp. [https://books.google.com/books?id=-yXMBgAAQBAJ&pg=PA80 80-81] (German). See also [[Sal Khan]]: [https://www.khanacademy.org/math/ap-calculus-ab/limits-from-equations-ab/squeeze-theorem-ab/v/proof-lim-sin-x-x ''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 {{mvar|x}} close enough to 0. The correctness of which for positive {{mvar|x}} can be seen by simple geometric reasoning (see drawing) that can be extended to negative {{mvar|x}} 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 {{mvar|x}} close enough to 0. This can be derived by replacing {{math|sin ''x''}} 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|thumb|upright=1.5|right]]

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 {{math|sec ''θ''}} and the arc on the [[unit circle]] has length&nbsp;{{math|Δ''θ''}}.  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 {{math|tan(''θ'' + Δ''θ'') − tan ''θ''}}, and the height is&nbsp;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&nbsp;{{math|Δ''θ'' > 0}}, and the inequalities are reversed if&nbsp;{{math|Δ''θ'' < 0}}.  Since the first and third expressions approach {{math|sec<sup>2</sup>''θ''}} as {{math|Δ''θ'' → 0}}, 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>{{cite book|chapter=Chapter 15.2 Limits and Continuity| pages=909–910|title=Multivariable Calculus|year=2008|last1=Stewart|first1=James| author-link1=James Stewart (mathematician)| edition=6th|isbn=978-0495011637}}</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 ==
=== Notes===
{{notelist}}

=== References ===
<references />

== External links ==
* {{MathWorld |title=Squeeze Theorem |urlname=SqueezeTheorem}}
* [http://demonstrations.wolfram.com/SqueezeTheorem/ Squeeze Theorem] by Bruce Atwood (Beloit College) after work by, Selwyn Hollis (Armstrong Atlantic State University), the [[Wolfram Demonstrations Project]].
* [https://proofwiki.org/wiki/Squeeze_Theorem Squeeze Theorem] on ProofWiki.

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