Solutions

Pictorially it means that $f(x)$ is bounded above and below. By definition, there exists a $\delta$ such that

$$0<|x-a|<\delta \Rightarrow |f(x)-l|<1.$$

Let $M=\max(|l-1|,|l+1|)$. We have

$$-M\leq |l-1|\leq l-1<f(x)<l+1\leq |l+1|\leq M.$$

Hence, $|f(x)|<M$.