How do I solve delta epsilon proofs I need to give an $ϵ,δ$ proof that $x^3$ is continuous at $x=-2$. I got that $|x^3+8|=|x+2||x^2-2x+4|$, and I need to assume $|x−2|<δ$. How can I choose $δ$ in terms of $ϵ$ so that $|x+2||x^2-2x+4|<ϵ$ ?
 A: A nice trick for problems like these is to simplify the expression for $|f(x) - f(x_0)|$ using inequalities.  Keep in mind that we're looking for any $\delta$ such that $|x+2| < \delta$ guarantees that $|f(x) - f(-2)| < \epsilon$; it doesn't have to be the best (i.e. largest) possible $\delta$.
To start, let's say that the $\delta$ we pick will definitely be smaller than $1$.  If that's the case, then we can guarantee that $-3 < x < -1$.  So, for any $x$ we consider it will be true that
$$
|x^2 - 2x + 4| \leq |x|^2 + 2|x| + 4 \leq 3^2 + 2 \cdot 3 + 4 = 19.
$$
That is, for any $x$ in the range $(-3,-1)$, we have
$$
|f(x) - f(-2)| = |x+2|\cdot |x^2 - 2x + 4| \leq 19 |x+2|.
$$
With that, we can simplify our task to the following: find a $\delta_1$ (in terms of $\epsilon$) such that if $|x + 2| < \delta_1$, then $19|x+2| < \epsilon$.  Having done this, we can take our $\delta$ to be $\min\{\delta_1,1\}$, so that if $|x+2| < \delta$ then we have both $|x+2| < 1$ and $|x+2| < \delta_1$.  So, we'll be able to say that if $|x+2|<\delta$, we have
$$
|f(x) - f(-2)| = |x+2|\cdot |x^2 - 2x + 4| \leq 19 |x+2| < \epsilon
$$
so that indeed, $|x+2| < \delta$ implies that $|f(x) - f(-2)| < \epsilon$.
A: Assuming that $|x+2|<\delta$ then $-4-\delta <x-2< -4+\delta$ so $(-4+\delta)^2 <|x-2|^2 <(4+\delta)^2<M$ (you can assume that you're taking $\delta <1$)
So there exist a positive bound $M$ such that $|x^2-2x+4|<M$ 
Take $\delta=\frac{\epsilon}{M}$
You'll get $|x+2||x^2-2x+4|< \frac{\epsilon}{M}.M= \epsilon$
