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


]
In ], the '''squeeze theorem''' (known also as the '''pinching theorem''', the '''sandwich theorem''', the '''sandwich rule''' and sometimes the '''squeeze lemma''') is a ] regarding the ].
]


In ], 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''. In Italy, the theorem is also known as the ''theorem of carabinieri''.}}) is a ] regarding the ] of a ] that is bounded between two other functions.
The squeeze theorem is used in calculus and ]. 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 ]s ] and ] in an effort to compute ], and was formulated in modern terms by ].


The squeeze theorem is used in calculus and ], 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 ] and ] in an effort to compute ], and was formulated in modern terms by ].
In ], ], ], ], ], ] and ], the squeeze theorem is also known as the '''two ] theorem''', '''two ] theorem''', '''] theorem''', '''two ] 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 == == Statement ==
The squeeze theorem is formally stated as follows.<ref>{{cite book|last1=Sohrab|first1=Houshang H.|title=Basic Real Analysis| date=2003|publisher=]|isbn=978-1-4939-1840-9|page=104|edition=2nd|url=https://books.google.com/books?id=QnpqBQAAQBAJ&pg=PA104}}</ref>
The squeeze theorem is formally stated as follows.
{{math theorem|
<blockquote>
Let ''I'' be an ] having the point ''a'' as a limit point. Let ''f'', ''g'', and ''h'' be ] defined on ''I'', except possibly at ''a'' itself. Suppose that for every ''x'' in ''I'' not equal to ''a'', we have: Let {{mvar|I}} be an ] containing the point {{mvar|a}}. Let {{mvar|g}}, {{mvar|f}}, and {{mvar|h}} be ] 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 ] (respectively) of {{mvar|f}}.
: <math>g(x) \leq f(x) \leq h(x) \, </math>
* Here, {{mvar|a}} is ''not'' required to lie in the ] 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===
and also suppose that:
According to the above hypotheses we have, taking the ] 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>\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L. \, </math>


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

</blockquote>
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. ]


The proof for sequences is very similar, using the <math>\varepsilon</math>-definition of the limit of a sequence.
* The functions ''g'' and ''h'' are said to be ] (respectively) of ''f''.
* Here ''a'' is ''not'' required to lie in the ] 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 == == Examples ==
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=== First example === === First example ===


] ]


The limit The limit


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


cannot be determined through the limit law cannot be determined through the limit law


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


because because


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


does not exist. does not exist.
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However, by the definition of the ], However, by the definition of the ],


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


It follows that It follows that


: <math>-x^2 \le x^2 \sin(\tfrac{1}{x}) \le x^2 \, </math> <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(\tfrac{1}{x})</math> must also be 0. 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 === === 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) \\
\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 \\
\Rightarrow & \sin x & \leq & x & \leq & \frac{\sin x}{\cos x} \\
\Rightarrow & \frac{\cos x}{\sin x} & \leq & \frac{1}{x} & \leq & \frac{1}{\sin x} \\
\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 Probably the best-known examples of finding a limit by squeezing are the proofs of the equalities
<math display="block">

: <math>
\begin{align} \begin{align}
& \lim_{x\to 0} \frac{\sin x}{x} =1, \\ & \lim_{x\to 0} \frac{\sin x}{x} =1, \\
Line 67: Line 110:
</math> </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. (German). See also ]: (video, ])</ref>
The first follows by means of the squeeze theorem from the fact that


: <math> \cos x < \frac{\sin x}{x} < 1 </math> <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
for ''x'' close enough, but not equal to 0.

<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. 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.
Line 78: Line 124:


It is possible to show that It is possible to show that
<math display="block"> \frac{d}{d\theta} \tan\theta = \sec^2\theta </math>

: <math> \frac{d}{d\theta} \tan\theta = \sec^2\theta </math>

by squeezing, as follows. by squeezing, as follows.


] ]


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


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


since the radius is sec&nbsp;''&theta;'' and the arc on the ] has length&nbsp;Δ''&theta;''. Similarly the area of the larger of the two shaded sectors is since the radius is {{math|sec ''θ''}} and the arc on the ] has length&nbsp;{{math|Δ''θ''}}. Similarly, the area of the larger of the two shaded sectors is


: <math> \frac{\sec^2(\theta + \Delta\theta)\,\Delta\theta}{2}. </math> <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&nbsp;tan(''&theta;''&nbsp;+&nbsp;Δ''&theta;'')&nbsp;&minus;&nbsp;tan(''&theta;''), and the height is&nbsp;1. The area of the triangle is therefore 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> \frac{\tan(\theta + \Delta\theta) - \tan(\theta)}{2}. </math> <math display="block"> \frac{\tan(\theta + \Delta\theta) - \tan\theta}{2}. </math>


From the inequalities From the inequalities


: <math> \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> <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 we deduce that


: <math> \sec^2\theta \le \frac{\tan(\theta + \Delta\theta) - \tan(\theta)}{\Delta\theta} \le \sec^2(\theta + \Delta\theta),</math> <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;Δ''&theta;''&nbsp;>&nbsp;0, and the inequalities are reversed if&nbsp;Δ''&theta;''&nbsp;<&nbsp;0. Since the first and third expressions approach sec<sup>2</sup>''&theta;'' as Δ''&theta;''&nbsp;&nbsp;0, and the middle expression approaches (''d''/''d&theta;'')&nbsp;tan&nbsp;''&theta;'', the desired result follows. 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 === === 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|authorlink1=James_Stewart_(mathematician)|edition=6th|isbn=0495011630}}</ref> 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>\lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2}</math> <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 cannot be found by taking any number of limits along paths that pass through the point, but since


<math display="block">\begin{array}{rccccc}
: <math>0 \leq \frac{x^2}{x^2+y^2} \leq 1</math>
: <math>-\left | y \right \vert \leq y \leq \left | y \right \vert </math> & 0 & \leq & \displaystyle \frac{x^2}{x^2+y^2} & \leq & 1 \\
: <math>-\left | y \right \vert \leq \frac{x^2 y}{x^2+y^2} \leq \left | y \right \vert </math> -|y| \leq y \leq |y| \implies & -|y| & \leq & \displaystyle \frac{x^2 y}{x^2+y^2} & \leq & |y| \\
{
: <math>\lim_{(x,y) \to (0, 0)} -\left | y \right \vert = 0</math>
: <math>\lim_{(x,y) \to (0, 0)} \left |y \right \vert = 0</math> {\displaystyle \lim_{(x,y) \to (0, 0)} -|y| = 0} \atop
: <math>0 \leq \lim_{(x,y) \to (0, 0)} \frac{x^2 y}{x^2+y^2} \leq 0</math> {\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, therefore, by the squeeze theorem,


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


== References == == References ==
=== Notes===
{{refimprove|date=April 2010}}
{{notelist}}
* {{MathWorld |title=Squeezing Theorem |urlname=SqueezingTheorem}}

=== References ===
<references /> <references />


== External links == == External links ==
* {{MathWorld |title=Squeezing Theorem |urlname=SqueezingTheorem}}
* by Bruce Atwood (Beloit College) after work by, Selwyn Hollis (Armstrong Atlantic State University), the ]. * by Bruce Atwood (Beloit College) after work by, Selwyn Hollis (Armstrong Atlantic State University), the ].
* on Proofs.wiki. * on ProofWiki.

{{Portal bar|Mathematics}}


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] ]
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Latest revision as of 09:32, 9 July 2024

Method for finding limits in calculus
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"Sandwich theorem" redirects here. For the result in measure theory, see Ham sandwich theorem. For Sandwich theory (physics), see Sandwich theory.
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 names) 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 π, and was formulated in modern terms by Carl Friedrich Gauss.

Statement

The squeeze theorem is formally stated as follows.

Theorem —  Let I be an interval containing the point a. Let g, f, 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 ) f ( x ) h ( x ) {\displaystyle g(x)\leq f(x)\leq h(x)} and also suppose that lim x a g ( x ) = lim x a h ( x ) = L . {\displaystyle \lim _{x\to a}g(x)=\lim _{x\to a}h(x)=L.} Then lim x a f ( x ) = L . {\displaystyle \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 → ∞.

This theorem is also valid for sequences. Let (an), (cn) be two sequences converging to ℓ, and (bn) a sequence. If n N , N N {\displaystyle \forall n\geq N,N\in \mathbb {N} } we have anbncn, then (bn) also converges to ℓ.

Proof

According to the above hypotheses we have, taking the limit inferior and superior: L = lim x a g ( x ) lim inf x a f ( x ) lim sup x a f ( x ) lim x a h ( x ) = L , {\displaystyle 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.

A direct 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 | x a | < δ , {\displaystyle |x-a|<\delta ,} we have | f ( x ) L | < ε . {\displaystyle |f(x)-L|<\varepsilon .} Symbolically,

ε > 0 , δ > 0 : x , ( | x a | < δ   | f ( x ) L | < ε ) . {\displaystyle \forall \varepsilon >0,\exists \delta >0:\forall x,(|x-a|<\delta \ \Rightarrow |f(x)-L|<\varepsilon ).}

As

lim x a g ( x ) = L {\displaystyle \lim _{x\to a}g(x)=L}

means that

ε > 0 ,   δ 1 > 0 : x   ( | x a | < δ 1     | g ( x ) L | < ε ) . {\displaystyle \forall \varepsilon >0,\exists \ \delta _{1}>0:\forall x\ (|x-a|<\delta _{1}\ \Rightarrow \ |g(x)-L|<\varepsilon ).} (1)

and lim x a h ( x ) = L {\displaystyle \lim _{x\to a}h(x)=L}

means that

ε > 0 ,   δ 2 > 0 : x   ( | x a | < δ 2     | h ( x ) L | < ε ) , {\displaystyle \forall \varepsilon >0,\exists \ \delta _{2}>0:\forall x\ (|x-a|<\delta _{2}\ \Rightarrow \ |h(x)-L|<\varepsilon ),} (2)

then we have

g ( x ) f ( x ) h ( x ) {\displaystyle g(x)\leq f(x)\leq h(x)} g ( x ) L f ( x ) L h ( x ) L {\displaystyle g(x)-L\leq f(x)-L\leq h(x)-L}

We can choose δ := min { δ 1 , δ 2 } {\displaystyle \delta :=\min \left\{\delta _{1},\delta _{2}\right\}} . Then, if | x a | < δ {\displaystyle |x-a|<\delta } , combining (1) and (2), we have

ε < g ( x ) L f ( x ) L h ( x ) L   < ε , {\displaystyle -\varepsilon <g(x)-L\leq f(x)-L\leq h(x)-L\ <\varepsilon ,} ε < f ( x ) L < ε , {\displaystyle -\varepsilon <f(x)-L<\varepsilon ,}

which completes the proof. Q.E.D

The proof for sequences is very similar, using the ε {\displaystyle \varepsilon } -definition of the limit of a sequence.

Examples

First example

x 2 sin ( 1 x ) {\displaystyle x^{2}\sin \left({\tfrac {1}{x}}\right)} being squeezed in the limit as x goes to 0

The limit

lim x 0 x 2 sin ( 1 x ) {\displaystyle \lim _{x\to 0}x^{2}\sin \left({\tfrac {1}{x}}\right)}

cannot be determined through the limit law

lim x a ( f ( x ) g ( x ) ) = lim x a f ( x ) lim x a g ( x ) , {\displaystyle \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 0 sin ( 1 x ) {\displaystyle \lim _{x\to 0}\sin \left({\tfrac {1}{x}}\right)}

does not exist.

However, by the definition of the sine function,

1 sin ( 1 x ) 1. {\displaystyle -1\leq \sin \left({\tfrac {1}{x}}\right)\leq 1.}

It follows that

x 2 x 2 sin ( 1 x ) x 2 {\displaystyle -x^{2}\leq x^{2}\sin \left({\tfrac {1}{x}}\right)\leq x^{2}}

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

Second example

Comparing areas:
A ( A D B ) A ( sector  A D B ) A ( A D F ) 1 2 sin x 1 x 2 π π 1 2 tan x 1 sin x x sin x cos x cos x sin x 1 x 1 sin x cos x sin x x 1 {\displaystyle {\begin{array}{cccccc}&A(\triangle ADB)&\leq &A({\text{sector }}ADB)&\leq &A(\triangle ADF)\\\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\\\Rightarrow &\sin x&\leq &x&\leq &{\frac {\sin x}{\cos x}}\\\Rightarrow &{\frac {\cos x}{\sin x}}&\leq &{\frac {1}{x}}&\leq &{\frac {1}{\sin x}}\\\Rightarrow &\cos x&\leq &{\frac {\sin x}{x}}&\leq &1\end{array}}}

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

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

cos x sin x x 1 {\displaystyle \cos x\leq {\frac {\sin x}{x}}\leq 1}

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

0 1 cos x x x {\displaystyle 0\leq {\frac {1-\cos x}{x}}\leq x} for x close enough to 0. This can be derived by replacing sin x in the earlier fact by 1 cos 2 x {\textstyle {\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 d d θ tan θ = sec 2 θ {\displaystyle {\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

sec 2 θ Δ θ 2 , {\displaystyle {\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

sec 2 ( θ + Δ θ ) Δ θ 2 . {\displaystyle {\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

tan ( θ + Δ θ ) tan θ 2 . {\displaystyle {\frac {\tan(\theta +\Delta \theta )-\tan \theta }{2}}.}

From the inequalities

sec 2 θ Δ θ 2 tan ( θ + Δ θ ) tan θ 2 sec 2 ( θ + Δ θ ) Δ θ 2 {\displaystyle {\frac {\sec ^{2}\theta \,\Delta \theta }{2}}\leq {\frac {\tan(\theta +\Delta \theta )-\tan \theta }{2}}\leq {\frac {\sec ^{2}(\theta +\Delta \theta )\,\Delta \theta }{2}}}

we deduce that

sec 2 θ tan ( θ + Δ θ ) tan θ Δ θ sec 2 ( θ + Δ θ ) , {\displaystyle \sec ^{2}\theta \leq {\frac {\tan(\theta +\Delta \theta )-\tan \theta }{\Delta \theta }}\leq \sec ^{2}(\theta +\Delta \theta ),}

provided Δθ > 0, and the inequalities are reversed if Δθ < 0. Since the first and third expressions approach secθ as Δθ → 0, and the middle expression approaches d d θ tan θ , {\displaystyle {\tfrac {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.

lim ( x , y ) ( 0 , 0 ) x 2 y x 2 + y 2 {\displaystyle \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 x 2 x 2 + y 2 1 | y | y | y | | y | x 2 y x 2 + y 2 | y | lim ( x , y ) ( 0 , 0 ) | y | = 0 lim ( x , y ) ( 0 , 0 )       | y | = 0 0 lim ( x , y ) ( 0 , 0 ) x 2 y x 2 + y 2 0 {\displaystyle {\begin{array}{rccccc}&0&\leq &\displaystyle {\frac {x^{2}}{x^{2}+y^{2}}}&\leq &1\\-|y|\leq y\leq |y|\implies &-|y|&\leq &\displaystyle {\frac {x^{2}y}{x^{2}+y^{2}}}&\leq &|y|\\{{\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}}}

therefore, by the squeeze theorem,

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

References

Notes

  1. Also known as the pinching theorem, the sandwich rule, the police theorem, the between theorem and sometimes the squeeze lemma. In Italy, the theorem is also known as the theorem of carabinieri.

References

  1. Sohrab, Houshang H. (2003). Basic Real Analysis (2nd ed.). Birkhäuser. p. 104. ISBN 978-1-4939-1840-9.
  2. Selim G. Krejn, V.N. Uschakowa: Vorstufe zur höheren Mathematik. Springer, 2013, ISBN 9783322986283, pp. 80-81 (German). See also Sal Khan: Proof: limit of (sin x)/x at x=0 (video, Khan Academy)
  3. Stewart, James (2008). "Chapter 15.2 Limits and Continuity". Multivariable Calculus (6th ed.). pp. 909–910. ISBN 978-0495011637.

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