Find a positive number $\delta<2$ such that $|x−2| < \delta \implies |x^2−4| < 1$ I have to find a positive number $\delta<2$  such that $|x−2| < \delta \implies |x^2−4| < 1$. 
I know that $ \delta =\frac{1}{|x+2|} $ has this behaviour, but it is not guaranteed for it to be less than two. I don't know what else to do. Thank you!
 A: The choice $\frac{1}{|x+2|}$ is inappropriate for a more fundamental reason. We want a $\delta$ such that for any $x$ such that $|x-2|\lt \delta$, we have $|x^2-4|\lt 1$. So in particular $\delta$ must not depend of $x$.  We now poceed to find an appropriate $\delta$. 
Note that $|x^2-4|=|x-2||x+2|$. We can make $|x-2|$ small by choosing $\delta$ small, but the $|x+2|$ term could spoil things. 
Suppose however that we make $\delta=\frac{1}{5}$. If $|x-2|\lt \delta$, then
$2-\frac{1}{5}\lt x\lt 2+\frac{1}{5}$. Thus $x+2$ is positive and less than $5$. It follows that $|x+2|\lt 5$, and therefore if $|x-2|\lt \delta$ we have
$$|x-2||x+2|\lt \frac{1}{5}\cdot 5=1.$$
A: Observe that : $|x^2 - 4| = |x-2||x+2| < \delta|x+2| < \delta\left(|x - 2| + 4\right) < \delta(\delta + 4) = \delta^2 + 4\delta$. Thus all we need is to solve for $\delta$:
$\delta^2 + 4\delta < 1 \iff (\delta + 2)^2 < 5 \iff \delta + 2 < \sqrt{5} \iff \delta < \sqrt{5} - 2$. For example we can take: $\delta = \dfrac{\sqrt{5} - 2}{2}$
