Prove $f(x) = x^3$ is continuous at $ x=2.$ I need to give an $\epsilon, \delta$ 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 $|x-2| < \delta$. How can I choose $\delta$ in terms of $\epsilon$ so that  $|x-2| |x^2 +2x +4| < \epsilon$ ?
We learned the "adding in zero" trick, but that gets me $\mid x^2 +2x+4 \mid = \mid x^2 - 4x + 6x + 4 \mid \leq \mid (x-2)^2\mid + \mid 6x \mid$. Is that the right step to take?
We haven't learned about bounding the function yet. In our in-class example, we learned how to make the terms cancel nicely to constants using the "adding in zero". Is that not possible here?

Let $\epsilon > 0$ and $\delta < \min{1, \epsilon/19}$. Let $x$ be within $\delta$ of $2$. Observe that
$$
|x^3 - 8 |  = |x-2||x^2+2x+4| = |x-2| |(x-2)^2 + 6x| = |x-2| |(x-2)^2 + 6(x-2) + 12|.
$$
Applying the triangle inequality to the final absolute value, we find that
$$
|x^3 - 8 | \leq |x-2|^3 + 6 |x-2|^2 + 12 |x-2| < \delta^3 + 6 \delta ^2 + 12 \delta. 
$$
Since $\delta < 1$, $\delta^n < \delta$ for all $n$. Thus
$$
|x^3 - 8 | < \delta^3 + 6 \delta ^2 + 12 \delta < 19 \delta \leq \epsilon.
$$
So the function is continuous at $2$, as desired.
