Find delta with a given epsilon for $\lim_{x\to-2}x^3 = - 8$

Question

Here is the problem. If

$$\lim_{x\to-2}x^3 = - 8$$

then find $\delta$ to go with $\varepsilon = 1/5 = 0.2$.
Is $\delta = -2$?

Answer 1

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.