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Thread: Epsilon-Delta Limits for a Problem

  1. #1 Epsilon-Delta Limits for a Problem 
    Member epidecus's Avatar
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    Jan 2013
    I was curious to see if I could manage differentiating the hyperbolic sine using the difference quotient. The algebra has been easy, and I've broken it down to the limiting coefficients. Problem is... I realize I don't know how to do epsilon-delta limits! (except for very trivial ones) Before we get there, here's my derivative...


    Skipping the few intermediate steps...

    I know I'm shooting for the hyperbolic cosine , so now it's just a matter of showing that those limits are both 1. And the only way to do this right is through epsilon-delta proofs.

    When I started learning single-variable calculus ahead of time, I sort of skipped over doing the actual proofs and instead just assumed their expected values for the expected result. Though I understood the concept behind it, I pretty much went on to learn differentiation and integration without a clue as to how to do these proofs (why? I don't know!). Now I've stumbled back upon this gap, and I'm hoping that a walkthrough with this particular problem will help me to understand them. Let's start with the first coefficient...

    Show that



    I know I have to relate epsilon and delta somehow, but I'm not sure where to even start. Hints/suggestions?
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  2. #2  
    Member epidecus's Avatar
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    Jan 2013
    Actually, I might have it... (editing)

    UPDATE: I'm simply mirroring the example from #4 >here<.

    (I'll use in place of to make the formulae a bit lighter)

    Proving the limit

    Show that: IF , THEN


    Multiplying both sides by ...

    Now, is derived, and I could relate epsilon and delta. However, the right side is still in terms of , meaning I'll have to simplify the problem. Because the limit deals with being close to , we can make the restriction , so that .

    With the original inequality , the denominator of the right side is a globally increasing function. So when is at a maximum the quotient as a whole will be at a minimum.

    The maximum of the denominator function is , therefore...

    Then we are left with two inequalities, and

    Given , let . (so that both inequalities are satisfied)

    I assume I now have to work each case and put it into the form , which should complete the proof. Am I on the right track?
    Last edited by epidecus; 05-27-2013 at 01:19 AM.
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