Proof the converse of theorem is false

Question

Q: If limit of $f(x)$, as $x$ approaches $c$, is $L$ Then limit of absolute value $f(x)$, as $x$ approaches $c$, is absolute value $L$ How do I prove the converse is false with precise definition of limit

Any help is appreciated

Answer 1

$$f(x)=\begin{cases} 1 & x\in\mathbb{Q}\\ -1 & x\in\mathbb{R\setminus\mathbb{Q}} \end{cases}$$

$\lim_{x\to0}|f(x)|=1$, obviously. Now show that, for any value of $a$ and any neighborhood $(-\delta,\delta)$ of $0$, there is an $x$ in that neighborhood such that $|f(x)-a|>0.5$.