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Epsilons and Deltas and Such

  1. Sep 12, 2009 #1
    Okay Then! :smile: I am going to start with a simple problem here:

    Given some function, a limit L, an xo, and some [itex]\epsilon[/itex]:

    a) Find an open interval on which the inequality [itex]|f(x)-L|<\epsilon[/itex] holds. Then b) give a value for [itex]\delta>0[/itex] such that
    for all x satisfying 0 < |x - x0| < [itex]\delta\Rightarrow |f(x)-L|<\epsilon[/itex].

    f(x)=x+1
    L = 5
    xo=4
    [itex]\epsilon[/itex]=0.01

    a) To find an interval on which [itex]|f(x)-L|<\epsilon[/itex] holds, I simply solve the inequality:

    [tex]|f(x)-L|<\epsilon[/tex]

    [tex]-\epsilon<f(x)-L<\epsilon[/tex]

    [tex]-\epsilon<x+1-5<\epsilon[/tex]

    [tex]-\epsilon<x-4<\epsilon[/tex]

    [tex]4-\epsilon<x<4+\epsilon[/tex]

    [tex]3.99<x<4.01[/tex]

    So there is my open interval, (3.99, 4.01), on which [itex]|f(x)-L|<\epsilon[/itex] holds.

    Now I know that for part (b), delta must be 0.01.

    But what how do we actually find [itex]\delta[/itex]? What are the mechanics of finding it.

    For part (a) I solved an inequality; what did I do for part (b) to find delta?

    Sorry if this is a little vague, I am not sure exactly how to word my question.
     
  2. jcsd
  3. Sep 12, 2009 #2
    A better question to ask is what is delta? Delta is how sufficiently close x must be to x_0 to guarantee that |f(x) -L| < epsilon will be satisfied. Thus, we want [itex]|x-x_0| < \delta[/itex] or [itex]x_0 - \delta < x < x_0 + \delta.[/itex] In this case, [itex]x_0 = 4[/itex] so we want a delta such that [itex]4 -\delta < x < 4 + \delta[/itex].

    Now compare this last set of inequalities to what you concluded in part a). How do we choose delta to ensure that |f(x) - L| is indeed less than epsilon?
     
  4. Sep 12, 2009 #3
    So since we know that we need [itex]|x-x_0| < \delta[/itex] (1), we also know xo=4 (2), and we have an inequality that says 3.99 < x < 4.01 (3), we can simply 'combine' (1), (2), and (3) to yield ....err something. I need a moment to think about it. But, I think I see it now.
     
  5. Sep 12, 2009 #4
    I am still a little lost here :redface: sorry. What is the next step?

    I know that 3.99<x<4.01. I also know that |x-xo|<[itex]\delta[/itex]. And xo=4.

    How do I combine the 3 into something meaningful to find [itex]\delta[/itex] ?

    Thanks
     
  6. Sep 12, 2009 #5
    Nevermind. I was forgetting to subtract 4 from ALL sides of the inequality.

    Using the above we have:

    3.99-4 < x-4 < 4.01-4

    -0.01<x-4< 0.01
    or
    |x-4|<0.01 = delta
     
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