squeeze rule
Let $f,g,h:\mathbb{N}\to \mathbb{R}$ be three sequences of real numbers such that
$$f(n)\le g(n)\le h(n)$$ 
for all $n$. If ${lim}_{n\to \mathrm{\infty}}f(n)$ and ${lim}_{n\to \mathrm{\infty}}h(n)$ exist and are equal, say to $a$, then ${lim}_{n\to \mathrm{\infty}}g(n)$ also exists and equals $a$.
The proof is fairly straightforward. Let $\u03f5$ be any real number $>0$. By hypothesis^{} there exist $M,N\in \mathbb{N}$ such that
$$ 
$$ 
Write $L=\mathrm{max}(M,N)$. For $n\ge L$ we have

•
if $g(n)\ge a$:
$$ 
•
else $$ and:
$$
So, for all $n\ge L$, we have $$, which is the desired conclusion^{}.
Squeeze rule for functions
Let $f,g,h:S\to \mathbb{R}$ be three realvalued functions on a neighbourhood $S$ of a real number $b$, such that
$$f(x)\le g(x)\le h(x)$$ 
for all $x\in S\{b\}$. If ${lim}_{x\to b}f(x)$ and ${lim}_{x\to b}h(x)$ exist and are equal, say to $a$, then ${lim}_{x\to b}g(x)$ also exists and equals $a$.
Again let $\u03f5$ be an arbitrary positive real number. Find positive reals $\alpha $ and $\beta $ such that
$$ 
$$ 
Write $\delta =\mathrm{min}(\alpha ,\beta )$. Now, for any $x$ such that $$, we have

•
if $g(x)\ge a$:
$$ 
•
else $$ and:
$$
and we are done.
Title  squeeze rule 

Canonical name  SqueezeRule 
Date of creation  20130322 13:46:31 
Last modified on  20130322 13:46:31 
Owner  Daume (40) 
Last modified by  Daume (40) 
Numerical id  4 
Author  Daume (40) 
Entry type  Theorem^{} 
Classification  msc 26A03 
Synonym  squeeze theorem 
Synonym  squeeze test 