1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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].

    L = 5

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







    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] ?

  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
    |x-4|<0.01 = delta
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook