Q:

Prove $f(x) =x^2$ is continuous at $x_0=2$ using the $\epsilon$-$\delta$ definition

We want to prove $f(x) =x^2$ is continuous at $x_0=2$ using the $\epsilon$-$\delta$ definition.

My attempt:

We want the function $f$ to satisfy the definition of continuity, meaning :
For all $\epsilon \gt 0$ there exists $\delta$ such that at $x_0=2$, $ \lvert x-2 \rvert \lt \delta \implies \lvert f(x) - 4 \rvert \lt \epsilon$
Notice that $f(x)$ is really just $x^2$. Substituting this in we get $ \lvert x^2 - 4 \rvert \lt \epsilon$. 
$ \lvert x^2 - 4 \rvert = \lvert x-2 \rvert \lvert x+2 \rvert \lt \epsilon$.
This is where I get stuck:

How do we recognize the correct choice of $\epsilon$ and $\delta$ at this point?

Thank you!

A:

Note that $|x+2|<5$ for $|x-2| < 1$. So lets take $|x-2| < 1$. Then you have
$$|x-2||x+2| < |x-2|5$$
If $|x-2| < \tfrac \epsilon5$ you get $|x-2|5 < \epsilon$. We want

$|x-2| < 1$
$|x-2| < \tfrac\epsilon5$

Thus we take $\delta = \min\{1,\tfrac\epsilon5\}$, because if $|x-2|<\min\{1,\tfrac\epsilon5\}$ then both of the above conditions are fulfilled.

