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Epsilon-delta definition and continuity at a point

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the following equation is continous using the epsilon-delta definition at y=-2

    2. Relevant equations

    [tex]\f(y)=\sqrt[3]{y+3}[/tex]

    3. The attempt at a solution

    so i got to a stage where;

    [tex]\frac{1}{\left|\left(\sqrt[3]{y+3}\right)^{2}+\sqrt[3]{y+3}+1\right|}\times\left|y-c\right|[/tex]

    but this is where i always get stuck on these things. I just don't know where to start with picking my delta. If anyone could explain the best way of going about it that would be very helpful.

    Thanks

    p.s. i didn't type out the whole extent of my solution so far. Hopefully its ok so far but let me know if that bit is wrong as well.
     
  2. jcsd
  3. Nov 12, 2009 #2

    Mark44

    Staff: Mentor

    I revised your LaTeX so that it would render as you wanted it to.
    I think I understand what you're doing, with the idea being that (a - b)(a2 + ab + b2) = a3 - b3. I tried a different approach that I think will work.

    I'm assuming you want to show that [tex]\lim_{y \rightarrow -2} f(y) - f(-2) = 0[/tex]

    This means you want to show that for any [itex]\epsilon[/itex] > 0 there is a number [itex]\delta[/itex] > 0 such that |f(y) - f(-2)| < [itex]\epsilon[/itex] when |x - (-2)| < [itex]\delta[/itex]
    So given such an [itex]\epsilon[/itex], we want
    [tex]|(y + 3)^{1/3} - 1| < \epsilon[/tex]
    [tex]-\epsilon + 1 < (y + 3)^{1/3} < \epsilon + 1[/tex]
    [tex](-\epsilon + 1)^3 < y + 3 < (\epsilon + 1)^3[/tex]
    [tex](-\epsilon + 1)^3 - 1 < y + 2 < (\epsilon + 1)^3 -1[/tex]

    All that's left is to figure out what you need to use for your [itex]\delta[/itex].
     
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