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Math Help - Formal definition of limits and proofs using them.

  1. #1
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    Formal definition of limits and proofs using them.

    I have a couple of problems due soon that I need help comprehending.

    Firstly, I need to prove the Squeeze Theorem using the formal (Epsilon-Delta) definition of limits. I've read the actual proof from Wikipedia, but I just can't wrap my head around it, it feels like they're skipping some sort've explination in certain parts.

    For example, on the Wiki page (Squeeze theorem - Wikipedia, the free encyclopedia) it gets to a point where they give , then jump down and say that this proves , but I don't see how they got from one step to the other.

    Then, while they prove the general case, they have and end up with . I understand the two steps they show in between, but I don't see how using those two steps results in the latter equation.



    Secondly, I need to prove that: lim x->infinity f(x) is equal to lim x->0+ f(1/x), not intuitively, but using:

    The definition of lim x-> infinity f(x) = L is: for all epsilon > 0 there exists an N > 0 such that x > N -> |f(x)-L| < epsilon.

    Preemptive thanks for the help!
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  2. #2
    Senior Member tukeywilliams's Avatar
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    If  0 < |x-a| < \delta then  |f(x)| < \epsilon which implies that  \lim_{x \to a} f(x) = 0 (using definition of limit).
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  3. #3
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    Oh duh, I totally get that part now. It's kinda hard to follow logic when they're bombarding you with different equations and function representations.
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  4. #4
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    \lim_{x\to a}f(x) = L and \lim_{x\to a}g(x) = L means:
    0<|x-a|<\delta_1 \implies |f(x)-L|<\epsilon.
    0<|x-a|<\delta_2 \implies |g(x)-L|<\epsilon
    Choose \delta = \min \{ \delta_1,\delta_2 \}.
    This will imply that,
    0<|x-a|<\delta \implies |f(x)-L|<\epsilon \mbox{ and }|g(x)-L|<\epsilon.
    Now argue that since,
    g(x)\leq h(x)\leq f(x)
    We must have that,
    |h(x) - L|<\epsilon \mbox{ for }0<|x-a|<\delta.
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  5. #5
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    I've officially solved the first question, I just had to sit there and stare at the proof until it finally clicked. Now to work on the other one.
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