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Math Help - 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 \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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