I revised your LaTeX so that it would render as you wanted it to.
jameswill1am said:
Homework Statement
Show that the following equation is continuous using the epsilon-delta definition at y=-2
Homework Equations
[tex]f(y)=\sqrt[3]{y+3}[/tex]
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.
I think I understand what you're doing, with the idea being that (a - b)(a
2 + ab + b
2) = a
3 - b
3. 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].