Answer
$\delta=\sqrt5-2$
Work Step by Step
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-2|\lt\delta\Rightarrow|f(x)-4|\lt1$$
Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $3$ and $5$ (which means $|f(x)-4|\lt1$), $x$ must be placed between $\sqrt3$ and $\sqrt5$.
- Calculate the distance from $2$ to $\sqrt3$ and $\sqrt5$: $2-\sqrt3\approx0.268$ and $\sqrt5-2\approx0.236$.
So we would go with the nearer endpoint, which is $\sqrt5$ and take $\delta$ to be $\sqrt5-2$ ( or any smaller positive number is fine).
Therefore, $$0\lt |x-2|\lt\sqrt5-2\Rightarrow|f(x)-4|\lt1$$
