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Given epsilon, find delta

  1. Oct 23, 2008 #1
    I don't know how yor format this, so:

    x3 + x2 +x +1

    The limit of that function = 0 as x approaches -1

    What's the greatest value of delta when epsilon = 0.1?


    This is what I tried to do:

    |x3 + x2 + x + 1| < 0.1
    -0.1 < x3 + x2 + x + 1 < 0.1
    -1.1 < x3 + x2 + x < -0.9



    In my calculator I plotted y = f(x), y = -1.1, and y = -0.9
    The intersect are at x = -1.379 and x = -1.326

    So I plugged them in this equation.

    0<|x + 1|<delta

    |-1.379 + 1| < 0.379
    |-1.326 + 1| < 0.326

    So I pick delta = 0.326 because it's smaller.



    Now can someone please tell me everything that went wrong and offer any advice on how I could understand this?
     
  2. jcsd
  3. Oct 23, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What exactly was your f(x)? I do exactly what you say you did and I do NOT get -1.379 and -1.376. For one thing, x= -1 is NOT between those values- they can't possibly be right!
     
  4. Oct 23, 2008 #3

    Mark44

    Staff: Mentor

    Your delta is way too big. f(-1.326) = -0.8992, and f(-0.674) = 0.474094. Here x is within .326 of -1 (-1 - .326 and -1 + .326), but the function values are not within .1 of zero.

    When you plotted f(x) on your calculator, did you use f(x) = x^3 + x^2 + x + 1? You also show x^3 + x^2 + x in one of your inequalities, so I'm not sure whether you might have used that one.

    You should factor x^3 + x^2 + x + 1 into (x + 1) times the other factor.

    You want |x^3 + x^2 + x + 1| < epsilon
    so |x + 1|* |something| < epsilon

    You can tie that inequality to one involving |x + 1| if you replace |something| above with a value that's guaranteed to be larger, for all values of x on some reasonable interval.

    As a hint, delta needs to be quite a bit smaller than epsilon, like some fraction of it.
     
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