# 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) 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) and:

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

and we are done.

Title squeeze rule SqueezeRule 2013-03-22 13:46:31 2013-03-22 13:46:31 Daume (40) Daume (40) 4 Daume (40) Theorem msc 26A03 squeeze theorem squeeze test