Find delta with a given epsilon for $\lim_{x\to-2}x^3 = - 8$ Here is the problem. If 
$$\lim_{x\to-2}x^3 = - 8$$
then find $\delta$ to go with $\varepsilon = 1/5 = 0.2$.
Is $\delta = -2$?
 A: Sometimes Calculus students are under the impression that in situations like this there is a unique $\delta$ that works for the given $\epsilon$ and that there is some miracle formula or computation for finding it.
This is not the case.  In certain situations there are obvious choices for $\delta$, in certain situations there are not.  In any case you are asking for some $\delta\gt 0$ (!!!) such that for all $x$ with $|x-(-2)|\lt\delta$ we have 
$|x^3-(-8)|\lt 0.2$.
Once you have found some $\delta\gt 0$ that does it, every smaller $\delta\gt 0$
will work as well.  
This means that you can guess some $\delta$ and check whether it works.
In this case this is not so difficult as $x^3$ increases if $x$ increases.
So you only have to check what happens if you plug $x=-2-\delta$ and $x=-2+\delta$ into $x^3$
and then for all $x$ with $|x-(-2)|$ you will get values of $x^3$ that fall between these two extremes.  
For an educated guess on $\delta$, draw a sketch.
This should be enough information to solve this problem.
