|
Squeeze rule for sequences
Let $f,g,h:\N\to\R$ be three sequences of real numbers such that $$f(n)\le g(n)\le h(n)$$ 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 $\e$ be any real number $>0$ . By hypothesis there exist $M,N\in\N$ such that $$|a-f(n)|<\e\text{ for all }n\ge M$$ $$|a-h(n)|<\e\text{ for all }n\ge N$$ Write $L=\max(M,N)$ . For $n\ge L$ we have
- if $g(n)\ge a$ : $$|g(n)-a|=g(n)-a\le h(n)-a<\e$$
- else $g(n)<a$ and: $$|g(n)-a|=a-g(n)\le a-f(n)<\e$$
So, for all $n\ge L$ , we have $|g(n)-a|<\e$ , which is the desired conclusion.
Squeeze rule for functions
Let $f,g,h:S\to\R$ be three real-valued 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 $\e$ be an arbitrary positive real number. Find positive reals $\alpha$ and $\beta$ such that $$|a-f(x)|<\e\text{ whenever }0<|b-x|<\alpha$$ $$|a-h(x)|<\e\text{ whenever }0<|b-x|<\beta$$ Write $\delta=\min(\alpha,\beta)$ . Now, for any $x$ such that $|b-x|<\delta$ , we have
- if $g(x)\ge a$ : $$|g(x)-a|=g(x)-a\le h(x)-a<\e$$
- else $g(x)<a$ and: $$|g(x)-a|=a-g(x)\le a-f(x)<\e$$
and we are done.
|
