Prove $f(x) = x^3$ is continuous at $ x=2.$

Question

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?

Answer 1

One thing you might do is say up front that you'll never choose $\delta > 1$. In that case, you know that $x \in [1,3]$, which lets you bound $x^2 + 2x + 4$ above by $19$.

You can continue along with $|x - 2| |x^2 + 2x + 4| < 19|x - 2| = 19\delta$. If you want this to be less than $\epsilon$, then you can choose $\delta = \min\{1, \epsilon/19\}$

Answer 2

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.