squeeze rule

Let $f,g,h:\mathbb{N}\to\mathbb{R}$ be three sequences of real numbers such that

f(n)\leq g(n)\leq h(n)

for all $n$ . If $\lim_{n\to\infty}f(n)$ and $\lim_{n\to\infty}h(n)$ exist and are equal, say to $a$ , then $\lim_{n\to\infty}g(n)$ also exists and equals $a$ .

The proof is fairly straightforward. Let $\epsilon$ be any real number $>0$ . By hypothesis there exist $M,N\in\mathbb{N}$ such that

|a-f(n)|<\epsilon\text{ for all }n\geq M

|a-h(n)|<\epsilon\text{ for all }n\geq N

Write $L=\max(M,N)$ . For $n\geq L$ we have

•

if $g(n)\geq a$ :

$|g(n)-a|=g(n)-a\leq h(n)-a<\epsilon$
•

else $g(n)<a$ and:

$|g(n)-a|=a-g(n)\leq a-f(n)<\epsilon$

So, for all $n\geq L$ , we have $|g(n)-a|<\epsilon$ , which is the desired conclusion.

Squeeze rule for functions

Let $f,g,h:S\to\mathbb{R}$ be three real-valued functions on a neighbourhood $S$ of a real number $b$ , such that

f(x)\leq g(x)\leq 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 $\epsilon$ be an arbitrary positive real number. Find positive reals $\alpha$ and $\beta$ such that

|a-f(x)|<\epsilon\text{ whenever }0<|b-x|<\alpha

|a-h(x)|<\epsilon\text{ whenever }0<|b-x|<\beta

Write $\delta=\min(\alpha,\beta)$ . Now, for any $x$ such that $|b-x|<\delta$ , we have

•

if $g(x)\geq a$ :

$|g(x)-a|=g(x)-a\leq h(x)-a<\epsilon$
•

else $g(x)<a$ and:

$|g(x)-a|=a-g(x)\leq a-f(x)<\epsilon$

and we are done.

Title	squeeze rule
Canonical name	SqueezeRule
Date of creation	2013-03-22 13:46:31
Last modified on	2013-03-22 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