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: Epsilon and delta

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    For the limit below, find values of δ that correspond to the ε values.
    symimage.gif

    2. Relevant equations
    epsilon = .5
    and
    epsilon = .05


    3. The attempt at a solution
    These kinds of problems do you have to use a graphing calculator to figure it out?
    for epsilon = .05
    |(9x + x - 3x^3)-7|<.05
    6.95<(9+x-3x^3)<7.05

    and I graph it, i get x = about -.75548, y=6.95
    x = -.73803, y=7.05

    i get |x-1|<0.2619

    but its incorrect
     
  2. jcsd
  3. Jan 31, 2010 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Algebra could be used, couldn't it? Isn't solving quadratic inequalities one of the things they teach in pre-calculus?



    (I haven't checked your arithmetic, I'm assuming it's right)

    Anyways, you are misunderstand something. You did discover* that the interval** (-7.55, -7.39) does have the property that, for every x in it, f(x) lies within the interval (6.95, 7.95).

    But that interval is not described by the inequality |x-1|<0.2619....

    *: Well, more precisely, you have some evidence to suggest it. To really be confident in it, you have to find some algebraic proof, or a deeper understanding of approximations and conic functions.
    **: Of course, you found a slightly larger interval, but the difference isn't really relevant.
     
    Last edited: Jan 31, 2010
  4. Jan 31, 2010 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Try again! First off, those values are nowhere near 1. That should have been a first hint. Secondly, those values do not yield anything close to 7. The values for 9+x-3x3 for x=-0.75548 and x=-0.73803 are 2.538 and 2.438.

    You made another mistake here. -0.75-1=-1.75, not 0.25.
     
  5. Jan 31, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Try considering three separate limits and make use of the triangle inequality to find a delta that'll work for a given epsilon. That way you won't have a hard cubic to solve.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook