Editing Squeeze theorem (section)

== 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.