Solutions

Pick any $\varepsilon>0$. We now formulate a process of finding a $\delta>0$ to prove that

$$\lim_{x\to a} f(x)=0$$

for $a\in[0,1]$ using the $\varepsilon$-$\delta$ definition. Consider $x$ such that

$$0<|x-a|<\delta_a=\min(|a|,|1-a|).$$

Either $\max(f(x))$ over this range is $0$, or $1/n$ for some $n$ such that $x\in A_n$. If it is the former, we have a $\delta$ that proves the limit. Otherwise, we check if

$$\frac{1}{n}<\varepsilon.$$

If so, the same $\delta$ also proves the limit. If not, we shrink the range and consider $x$ such that

$$0<|x-a|<\delta_b=\min(|y-a|:y\in A_n).$$

Notice that $f(x)<1/n$ over this range. At this point we ask ourselves the same questions. If the $\delta$ is not satisfactory, we shrink the range again, each time lowering $\max(f(x))$ until either $\max(f(x))=1/m<\varepsilon$ for some $m\in\mathbb{N}$ or $\max(f(x))=0$.