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Delta and Epsilon Proofs

  1. Feb 27, 2015 #1
    Given a function f(x), a point x0, and a positive number E (epsilon), write the limit then find delta>0 such that for all x 0< |x-x0| < delta -> |f(x)-L| < E
    f(x) = 3-2x, x0=3, E=.02
    Here is my attempt:
    Lim (3-2x) as x->3 = -3
    -.02 < |3-2x - 3| <.02
    -.02 < |-2x| < .02
    .01 > x > -.01
    -2.99 > x-3 > -3.01
    -2.99 > |x-3|
    Delta= -2.99
    Is this right? I'm really confused, any help would be greatly appreciated, even just an explanation. Thanks.
     
  2. jcsd
  3. Feb 27, 2015 #2

    jedishrfu

    Staff: Mentor

  4. Feb 27, 2015 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You started going off track when you wrote |f(x)-L|<E. Since L is -3 that becomes |(3-2x)-(-3)|=|3-2x+3|=|6-2x|<0.02. Try it again from there.
     
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