# Thread: Proving squeeze theorem using epsilon delta proof

1. ## Proving squeeze theorem using epsilon delta proof

Anyone know how I'd go about doing this?

3. ## 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

4. ## 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?

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### proof squeeze theorem epsilon delta

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