I have the following problem:
Find $\delta>0$ s.t. for every x for which it is true that $|x-2| < \delta$ it is also true that $|x^2-4|<0.1$.
Here's the solution which I don't quite get:
From $|x^2-4|=|x-2||x+2|\le |x-2|(|x|+2)$ we see that we have restrictions for $|x|$ and that's why we'll look at numbers in the open interval with center 2 and radius 1 i.e. we are looking for $0 < \delta \le 1$. It is clear that $1<x<3$ or $|x|<3$. From there we see that to solve the problem we have to find $\delta$ s.t. $5\delta<0.1$ so we set $\delta =\frac{0.1}{5}$.
Why do we look at the open interval with center 2 and radius 1, how did we came up with that?
Thanks in advance!
We want to prove that $|x-2|\bigl(|x|+2\bigr)$ is small when $|x-2|$ (which is the first factor) is small. Since we already know that the first factor is small, you have to be sure that the second factor doesn't get very large. A way of doing this is to impose that $|x-2|\leqslant1$, which means that $1\leqslant x\leqslant3$, which in turn implies that $|x|+2<5$.
Theres is nothing magical about the choice of $1$. It's just a natural choice for a positive number. Suppose that we decide that our first step is to impose that $|x-2|\leqslant2$ instead. Then the conclusion is that $|x|+2\leqslant6$, and so we pick $\delta<\frac{0.1}6$ (which is $\leqslant2$, of course).