Idea of the $\min$ function in proving limits

Question

Suppose you want to prove $|x - a||x + a| < \varepsilon$

You know

$$|x - a| < (2|a| + 1)$$ You need to prove

$$|x + a| < \frac{\varepsilon}{2|a| + 1}$$

So that

$$|x - a||x + a| < \varepsilon$$

Why does Michael Spivak do this:

He says you have to prove $|x + a| < \min\left(1, \dfrac{\varepsilon}{2|a| + 1}\right)$ in order to finally prove $|x + a||x - a| < \varepsilon$.

Why do we need the $\min$ function there?

Thanks!

Answer 1

It will try to guess what it is about, but I may be wrong.

First, the estimation $$ |x-a|=|x+a-2a|\le|x+a|+2|a|<1+2|a|\tag1 $$ is valid only for $|x+a|<1$. That's why he needs to prove $$ |x+a|<\frac{\epsilon}{1+2|a|}\qquad\text{and}\qquad |x+a|<1 \tag2 $$ in order to be able to combine $(1)$ and $(2)$. The condition $(2)$ can be equivalently written as $$ |x+a|<\min\left\{1,\frac{\epsilon}{1+2|a|}\right\}. $$