Check my proof for this limit relation please

Question

Consider two functions $f, g:\mathbb{R} \rightarrow \mathbb{R} $ and $a\in \mathbb{R} $. We know that $\lim_{x \to a} f(x) =0$ and $g$ is bounded.
Is it true that $\lim_{x \to a} f(x) \cdot g(x) =0$?
My attempt :Yes, it is.
Since $g$ is bounded $\exists M>0$ so that $-M\le g(x) \le M$.
If we multiply this relation by $f(x) $ and apply the squeeze theorem we get that $\lim_{x \to a} f(x) \cdot g(x) =0$.

Answer 1

Be careful when you say multiply this relation by $f(x)$, we can't say $-M f(x) \le f(x) g(x) \le M f(x)$ because $f(x)$ could have been negative.

But yup, if we have $|g(x)| \le M$ and we can multiply by $|f(x)|$ and obtain $$0\le|f(x)g(x)|\le M|f(x)|$$ and now we can use squeeze theorem.