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Math Help - Proving Limits of a

  1. #1
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    Proving Limits of a

     \lim_{x \rightarrow a} x = a
    this is my scratch work *

     0< \left | x-a \right | < \delta

      \left | x-a \right | < \epsilon

    I let
      \delta = \epsilon
    Im not sure if this is right it seems to simple. I am confused by this because i have only worked with numbers.
    Help me please!
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  2. #2
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    Re: Proving Limits of a

    Quote Originally Posted by mathisfun26 View Post
     \lim_{x \rightarrow a} x = a
    this is my scratch work *

     0< \left | x-a \right | < \delta

      \left | x-a \right | < \epsilon

    I let
      \delta = \epsilon

    It may be too simple, but it is correct.
    Thanks from mathisfun26
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  3. #3
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    Re: Proving Limits of a

    Thank you!
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  4. #4
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    Re: Proving Limits of a

    It is simple, and I'll bet your thinking is right, but I have a few corrections.

    There's a good paper on these epsilon-delta proofs here at MHF - it's one of the four "sticky" threads in this forum (the Calculus forum).

    To prove \lim_{x\to{a}}f(x)=L:

    For the scratch work, you start with | f(x) - L | < \epsilon and work backwards to | x-a | < some function of \epsilon. That tells you what you need to use for \delta. In your case, this involves no steps (!), as you start with | x-a | < \epsilon and it is already in the right format. So you set \delta=\epsilon, which is what you said. I'm not sure why you had \delta earlier.

    Now you need to write the proof. It looks like you might have skipped this step - which would be very important on an exam.....

    I'm also a little confused as to what you mean by you've only worked with numbers.

    - Hollywood
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  5. #5
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    Re: Proving Limits of a

    what hollywood is saying is: to prove a limit exists, GIVEN epsilon, you have to somehow produce delta. you can't "start with delta".

    a proper proof (which hides the "scratch work", but that's ok, if sometimes mysterious) would go something like:

    let ε > 0. then for δ = ε,

    we have: 0 < |x - a| < δ implies |x - a| < ε.

    that is, for f(x) = x, we have:

    \lim_{x \to a} f(x) = \lim_{x \to a} x = a. since this also equals f(a), this shows that the identity function f(x) = x, is continuous (which really shouldn't be that surprising...if x is near a, then ...x is near a!).
    Last edited by Deveno; January 11th 2013 at 07:04 PM.
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