# Proving squeeze theorem using epsilon delta proof

• Aug 21st 2012, 04:51 PM
Jerry99
Proving squeeze theorem using epsilon delta proof
Anyone know how I'd go about doing this?
• Aug 21st 2012, 05:20 PM
HallsofIvy
Re: Proving squeeze theorem using epsilon delta proof
• Aug 21st 2012, 05:52 PM
Jerry99
Re: Proving squeeze theorem using epsilon delta proof
I guess I'll just play with some sin functions with a limit of 1 see if I can prove it with epsilon delta. I'm just still a little confused with epsilon delta. Thanks for the tip
• Aug 21st 2012, 06:50 PM
HallsofIvy
Re: Proving squeeze theorem using epsilon delta proof
Suppose $\lim_{x\to a} f(x)= A$, $\lim_{x\to a} g(x)= A$ and $f(x)\le h(x)\le g(x)$ at least for x in some neighborhood of a. Then given any $\epsilon> 0$, because $\lim_{x\to a} f(x)= A$, there exist $\delta_1> 0$ such that if $|x- a|< \delta_1$, then $|f(x)- a|< \epsilon$. Because $\lim_{x\to a} g(x)= A$ there also exist $\delta_2> 0$ such that if $|x- a|< \delta_2$, then $|g(x)- a|< \epsilon$. Of course, it is true that if $f(x)\le h(x)\le g(x)$, then $f(x)- a\le h(x)- a\le g(x)- a$. But that means that if |x- a| is less than the smaller of $\delta_1$ and $\delta_2$ we must have $-\epsilon< f(x)-a \le h(x)- a\le g(x)- a< \epsilon$. Can you finish from that?