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Thread: Proving squeeze theorem using epsilon delta proof

  1. #1
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    Proving squeeze theorem using epsilon delta proof

    Anyone know how I'd go about doing this?
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  2. #2
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    Re: Proving squeeze theorem using epsilon delta proof

    Yes. Thank you for asking.
    Thanks from Plato and Jerry99
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  3. #3
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    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
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  4. #4
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    Re: Proving squeeze theorem using epsilon delta proof

    Suppose $\displaystyle \lim_{x\to a} f(x)= A$, $\displaystyle \lim_{x\to a} g(x)= A$ and $\displaystyle f(x)\le h(x)\le g(x)$ at least for x in some neighborhood of a. Then given any $\displaystyle \epsilon> 0$, because $\displaystyle \lim_{x\to a} f(x)= A$, there exist $\displaystyle \delta_1> 0$ such that if $\displaystyle |x- a|< \delta_1$, then $\displaystyle |f(x)- a|< \epsilon$. Because $\displaystyle \lim_{x\to a} g(x)= A$ there also exist $\displaystyle \delta_2> 0$ such that if $\displaystyle |x- a|< \delta_2$, then $\displaystyle |g(x)- a|< \epsilon$. Of course, it is true that if $\displaystyle f(x)\le h(x)\le g(x)$, then $\displaystyle f(x)- a\le h(x)- a\le g(x)- a$. But that means that if |x- a| is less than the smaller of $\displaystyle \delta_1$ and $\displaystyle \delta_2$ we must have $\displaystyle -\epsilon< f(x)-a \le h(x)- a\le g(x)- a< \epsilon$. Can you finish from that?
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