# Epsilon-delta definition and continuity at a point

1. Nov 12, 2009

### jameswill1am

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

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

3. The attempt at a solution

so i got to a stage where;

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

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. Nov 12, 2009

### 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 $$\lim_{y \rightarrow -2} f(y) - f(-2) = 0$$

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

All that's left is to figure out what you need to use for your $\delta$.
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