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Illustrate the definition of a limit

  1. Sep 9, 2011 #1
    1. The problem statement, all variables and given/known data
    For the limit

    lim
    x → 2 (x^3 − 3x + 5) = 7

    illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.)


    2. Relevant equations



    3. The attempt at a solution
    I've gotten x^3-3x=2.2, but I'm lost after that. My teacher gave a terrible description on what to do, and the book is of no help whatsoever.
     
  2. jcsd
  3. Sep 9, 2011 #2

    lanedance

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    so lets start by writing x=2+d, with \delta =|d| and substitute in
    |((2+d)^3 − 3(2+d) + 5-7| <0.2
     
  4. Sep 9, 2011 #3
    Alright, so now I've got
    d^3 + 6d^2 + 9 < 0.2
     
  5. Sep 9, 2011 #4

    lanedance

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    That doesn't look quite right, i would check your steps. Also you can't drop the absolute value sign
     
  6. Sep 9, 2011 #5
    |(2+d)^3 -6 - 3d - 2| < 0.2
    |d^3 +6d^2 + 12d + 8 - 6 - 3d - 2| < 0.2

    = |d^3 + 6d^2 + 9d| < 0.2
     
  7. Sep 9, 2011 #6
    I don't see at all how this is helpful, and cannot recall when this serves as a good approach.

    fillipeano-
    I don't see how it is easier to solve it for a specific [itex]\epsilon[/itex] rather than a general one, so let's try and guide you on the general one. Then we'll just substitute the values given.

    So let [itex]\epsilon[/itex] > 0.
    As usually, you need to start from the end.
    Let's say f(x) = x3 -3x + 5
    I can understand of course how you got to the equation you wrote, but it should actually be an inequality. We need to find a [itex]\delta[/itex], so that for each x for which |x-2|<[itex]\delta[/itex]| holds, |f(x) - 7|<[itex]\epsilon[/itex] also holds. (instead of saying [itex]\epsilon[/itex] you could say "0.2")

    So, let see how we can "control" the expression |f(x)-7|:

    |x3 -3x + 5 - 7| = |x3 - 3x-2|.
    Now we're stuck.
    However, we know that this has to do somehow with "|x-2|", because we want to use that fact that |x-2|<[itex]\delta[/itex]|.
    That should make you think about polynomial division. Do you know the technique?
    Then you could have |x3 - 3x-2| = |x-2| * |P(x)| < [itex]\delta[/itex] |P(x)|,
    where P(x) is some polynomial of second order which you might also be able to control. :)
    See if you could take it from there...
     
  8. Sep 9, 2011 #7

    lanedance

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    in this case it simplifies down to
    |d^3 + 6d^2 + 9d| = |d^2 + 6d+ 9||d| = (d+3)^2|d|<e

    though it is the same thing really whether you work with |x-2| or |d|, just preference
     
  9. Sep 9, 2011 #8
    But what is the justification of this substitution? One needs to show that |f(x) - 7| < [itex]\epsilon[/itex]. So why do you try manipulating |f(x+[itex]\delta[/itex]) - 7|?
     
  10. Sep 9, 2011 #9

    HallsofIvy

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    Because the whole problem is to find [itex]|\delta[/itex] so that [itex]f(2+ \delta)- 7|< \epsilon[/itex]. It is NOT "to show that [itex]|f(x)- 7|<\epsilon[/itex]" because that, in general, is not true.
     
  11. Sep 9, 2011 #10
    Of course, but wouldn't you then want to show that |[itex]f(2- \delta)- 7|< \epsilon[/itex] as well?
    Looks like a strange approach to me, but if everyone's approving... :-)
    (for the same suitable [itex]\delta[/itex] that is)
     
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