Q:

Quick question about $\epsilon -\delta$ proofs

There is one step in $\epsilon - \delta$ proofs that I hope somebody could bring clarity to for me. Say we wanted to show $\displaystyle \lim_{x \to 2} x^2 = 4 $. Somewhere along the proof we would have the inequality $|x-2| \cdot |x+2| < \epsilon$ emerging. We could here assume $\delta \leq 1$ and thus have $|x + 2|<5$. This then means $$|x-2|\cdot |x+2|  < |x-2| \cdot 5$$
And here is where I get a bit dubious. Normally, one takes the step to casually claim that the following must hold: $$|x-2| \cdot 5 < \epsilon$$
and from there on we find (one of) our $\delta = \epsilon / 5$. 
But how can one be so certain of the claim that $|x-2| \cdot |x+2| < \epsilon$ and $|x+2| < 5$ implies $|x-2|\cdot 5 < \epsilon$? We are essentially claiming that $ab < \epsilon$ and $b < c$ implies $ab < ac < \epsilon$. But of course this cannot be true, right? For example $4\cdot 5 < 21$ and $5 < 6$ implying $4\cdot 5 < 4 \cdot 6 < 21$ is non-sensical, so what is really going on?

A:

Think of it this way. 
You know the inequality $|x-2| \cdot |x+2| < |x-2| \cdot 5$ is true. You are trying to show that the inequality $|x-2| \cdot |x+2| < \epsilon$ is true. By the transitive property of inequality, you will have successfully shown that inequality to be true if you are able to show that $|x-2| \cdot 5 < \epsilon$. This is equivalent to showing that $|x-2| < \frac{\epsilon}{5}$. So by choosing $\delta = \text{min}\{1,\frac{\epsilon}{5}\}$ you are done.

