Proving squeeze theorem using epsilon delta proof

**Jerry99** · August 21st 2012, 03:51 PM

Anyone know how I'd go about doing this?

**HallsofIvy** · August 21st 2012, 04:20 PM

Yes. Thank you for asking.

**Jerry99** · August 21st 2012, 04:52 PM

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

**HallsofIvy** · August 21st 2012, 05:50 PM

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?

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