# 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 $\delta = -2$?

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You need to find $\delta$ small enough that if $x$ is within $\delta$ of -2, then $x^3$ is within 0.2 of -8—that is, between 7.8 and 8.2. $\delta=-2$ is not nearly small enough, because then $x$ could be anywhere between -4 and 0, and $x^3$ might be very different from -8. – MJD Sep 19 '12 at 6:50
HINT: The idea is that you need to find $\delta$ such that if $$\left| -2 - x \right| \leq \delta$$ then $$\left|-8 - x^3 \right| \leq \frac{1}{5}$$ – Deven Ware Sep 19 '12 at 6:51
After Deven's hint, I'll recite my comment: Is it possible that $\delta<0$? – Dennis Gulko Sep 19 '12 at 6:53
really? I plugged in x^3 = -7.8 and x^3 = -8.2 – dsta Sep 19 '12 at 6:54
@Dennis In the problem it does not say that δ has to be greater than 0. – dsta Sep 19 '12 at 6:54

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.
Plug $-2.01$ and $-1.99$ into $x^3$ and see what happens. – Stefan Geschke Sep 19 '12 at 9:33