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Definition of a limit

  1. Sep 11, 2013 #1
    1. The problem statement, all variables and given/known data
    lim (x - 4) = 2
    as x -> 6

    eps > 0
    delta > 0

    No eps given

    2. Relevant equations


    3. The attempt at a solution
    Not sure how to apply equation..
    Usually I would do

    I assumed you would do..

    6 - eps < x - 4 - 2 < 6 + eps
    12 - eps < x < 12 + eps

    So, it would be min, (12-eps,12+eps)

    But what do I do from here?
     
    Last edited: Sep 11, 2013
  2. jcsd
  3. Sep 11, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    You seem to be confused with the core concept of this question ( The limit definition ). Here's an easy way to remember the definition in symbols:

    $$\forall \epsilon > 0, \exists \delta(\epsilon) > 0 \space | \space 0 < |x-a| < \delta \Rightarrow |f(x) - L| < \epsilon$$

    It basically reads : Forall positive epsilon, there exists a positive delta in terms of epsilon such that we can make ##f(x)## as close to ##L## as we like. A more in depth explanation of what's going on is below.

    Considering ##0 < |x-a| < \delta##, we know that ##|x-a|## must always be positive, so we never actually consider what happens at ##x=a##, only what happens as we approach it. That's where ##|x-a| < \delta## comes into play.

    Expanding we get ##-\delta < x-a < \delta## and then ##x + \delta > a > x -\delta##.

    So there exists a ##\delta > 0## such that ##a## is bounded between ##x + \delta## and ##x -\delta## and we can only arbitrarily get close to ##a##.

    Using this, what does it say about ##|f(x) - L| < \epsilon##?

    Well first, lets consider ##-\epsilon < f(x) - L < \epsilon## which yields ##f(x) + \epsilon > L > f(x) - \epsilon##. So ##\forall \epsilon > 0## we can make ##f(x)## as close to ##L## as we like. How close do we need to be you might ask? Sufficiently close.

    What defines sufficiently close ( You probably hear this a lot )? Well, ##f(x)## varies according to the values of ##x##. How far away ##x## is from ##a## depends on ##\delta##. So the values of how far ##f(x)## is away from ##L## ( which change according to ##\epsilon## ) will also vary according to ##\delta##. Hence ##\epsilon## will change according to the ##\delta## we choose.

    So in conclusion we know we can choose a ##\delta(\epsilon)## as to make ##f(x)## as close to ##L## as we like.

    So as for your question, start with the ##|f(x) - L|## portion and massage it a bit to get it into a suitable form.
     
    Last edited: Sep 11, 2013
  4. Sep 11, 2013 #3
    I just want to add this picture to supplement Zondrina explanation.
    precise.gif
    .
     
  5. Sep 11, 2013 #4
    Thanks a lot to both of you! When I get back from class I will thoroughly read through all of it and hopefully able to conceptually understand it.

    EDIT: I'm an idiot..

    i forgot delta = eps in linear functions with a slope of one ;D
     
    Last edited: Sep 11, 2013
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