prove squeeze theorem Can anyone prove Squeeze theorem for me, which says:
(1):if $f(x)< g(x) < h(x)$ for all $x$ near $a$, and $\lim\limits_{x\to a}f(x) = \lim\limits_{x \to a}h(x) = L$ 
then $\lim\limits_{x \to a}g(x)=L$.
(2): If $f$ dominates $g$, that is, ($f(x) \geq g(x)$) near $x=a$ and $g(x) \to \infty$ as $x \to \infty$ then so does $f(x)$
Thank you.
 A: Hints:
For 1: choose $\delta_f$ and $\delta_h$ such that $|x - a| < \delta_f$ implies $|f(x) - L| < \epsilon$ and $|x - a| < \delta_h$ implies $|h(x) - L| < \epsilon$.  Let $\delta$ be the smaller of $\delta_f$ and $\delta_h$.  Now if $|x - a| < \delta$ then both $|f(x) - L|$ and $|h(x) - L|$ are less than $\epsilon$.  Use that, along with $f(x) < g(x) < h(x)$, and you will be able to show that $|g(x) - L| < \epsilon$.  This will prove that $g(x) \to L$ as $x \to a$.
For 2: If $f$ dominates $g$ near $a$ then I suspect you want to take the limit as $x \to a$ right?  I'll leave this one to you, if you just write down the definition of the limit for $g$ you should be able to turn that into the definition for a limit of $f$.
A: We have 
$$|g(x)-L|\leq \max(|f(x)-L|,|h(x)-L|)$$
Let $\epsilon>0$ then there's $\delta'>0$, if $|x-a|\leq \delta'$ we have $|f(x)-L|<\epsilon$ and there's $\delta''>0$, if $|x-a|\leq \delta''$ we have $|h(x)-L|<\epsilon$
Take $\delta=\min(\delta',\delta'')$ and if $|x-a|\leq \delta$ we have $|g(x)-L|<\epsilon$
For the second result, let $A>0$ then there's $\delta>0$, if $|x-a|\leq \delta$ we have $f(x)\geq g(x)>A$
A: If $$f<g<h \ \forall \ x\in \Bbb R$$
and $$\lim_{x\to a}f=\lim_{x\to a}h=L$$
We have $g$ between $f  \ and \ h$ so the $g$ must be $L\Big|_{x\to a}$ , otherwise $f<g<h$ would not hold true.
In other words $f<g<h$ squeeze the $g$ between both $f,h$  when $f$ and $g$ tend to $L$
