Answer
$\delta=0.39$
Work Step by Step
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-3|\lt\delta\Rightarrow|f(x)-4|\lt0.2$$
Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $3.8$ and $4.2$ (which means $|f(x)-4|\lt0.2$), $x$ must be placed between $2.61$ and $3.41$.
- Calculate the distance from $3$ to $2.61$ and $3.41$: $3-2.61=0.39$ and $3.41-3=0.41$.
So we would go with the nearer endpoint, which is $2.61$ and take $\delta$ to be $0.39$ ( or any smaller positive number is fine).
Therefore, $$0\lt |x-3|\lt0.39\Rightarrow|f(x)-4|\lt0.2$$
