I know that if $\lim_{x \to a} f(x)=l \implies \lim_{x \to a} |f|(x)=|l|$.
If we know that $\lim_{x \to a} |f|(x)=|l|$ then what can we say about $\lim_{x \to a} f(x)$ ?
Are there any implications? If so then what are they? Is the implication that $\lim_{x \to a} f(x)=l$ or $-l$ true or false?
In general, we cannot conclude from $\lim_{x\to a} |f(x)|=l$ that $\lim_{x\to a} f(x)$ even exists.
(See the answer of haqnatural)
However, there are still some implications.
First case: $\lim f(x)$ exists
Assume that $\lim f(x)$ exists. The absolute value is a continuous function. Hence the limit operation can be exchanged with taking the absolut value. We get
$$ |\lim_{x\to a} f(x)|=\lim_{x\to a}|f(x)|$$
Thus if the limit $\lim_{x\to a} f(x)$ exists, then the limit $\lim_{x\to a} |f(x)|$ is $+\lim f(x)$ or $-\lim f(x)$.
Second case: $\lim |f(x)|=0$
It is always true that $-|f(x)|\le f(x) \le |f(x)|$. Hence the limit of this equation implies
$$0=-\lim_{x\to a}|f(x)|\le \liminf_{x\to a}f(x) \le\limsup_{x\to a}f(x) \le \lim_{x\to a}|f(x)|=0.$$ Thus $\liminf_{x\to a}f(x) =\limsup_{x\to a}f(x)=0$, which implies $\lim_{x\to a}f(x)=0$.
As Counter example take $$\lim _{ n\rightarrow \infty }{ \left| { \left( -1 \right) }^{ n } \right| } =1$$ however $$\lim _{ n\rightarrow \infty }{ { \left( -1 \right) }^{ n } } $$ limit not exsist
Nothing can be concluded from $\lim\limits_{x \to a} |f|(x)=|l|$. Consider the function $$ f(x) = \begin{cases} l &\text{if } x \in \Bbb Q \\ -l &\text{if } x \in \Bbb R \setminus \Bbb Q. \end{cases} $$
$|f| \equiv l$ is continuous everywhere on $\Bbb R$. (In particular, at $a \in \Bbb R$, $\lim\limits_{x \to a} |f|(x)$ exists and is equal to $|l|$.) However, due to the density of (ir)rational numbers in $\Bbb R$, $f$ is discontinuous everywhere, so the limit $\lim\limits_{x \to a} |f|(x)$ doesn't exist for any $a \in \Bbb R$.
Hence the limit of $|f(x)|$ at $a \in \Bbb R$ doesn't imply that of $f(x)$ at $a$.