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!

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: $\delta = \dfrac{\sqrt{5} - 2}{2}$
