I once saw the notation $O(x) \to 0$ or $o(x) \to 0$ for $x\to 0$ but I don't really know why that makes sense since $O(x)$ is more like a class of functions. So I would interpret this as "if $f(x) = O(x)$, then $f(x) \to 0$ for $x\to 0$". Does this always hold? Can I conclude from $$\lim_{x\to 0}\frac{f(x)}{x} = 0 \implies \lim_{x \to 0}f(x) = 0$$ ?
EDIT: I was referring to the first answer of this post.
$\displaystyle \lim_{x\to 0}\frac{f(x)}{x} = 0 \implies \lim_{x \to 0}f(x) = 0$ is certainly true since
if $\left|\frac{f(x)}{x}\right| \lt \epsilon$ for all $x$ with $0 \lt |x| \lt \delta$
then $\left|{f(x)}\right| \lt |x| \epsilon \lt \epsilon$ for all $x$ with $0 \lt |x| \lt \min(1,\delta)$
so you will be arbitrarily close to $0$ as $x \to 0$
EDIT: I misunderstood the question and assumed the $O$ referred to big-O notation. The correct interpretation is not this one; therefore my answer is rendered irrelevant.
As you noted, $O(x)$ is a class of functions therefore
$$f(x) = O(x)$$ makes no sense assuming you meant that $f(x)$ is a function.
One can write, however, $f(x) \in O(x)$.
Remember that one writes $f(x) \in O(g(x))$ if, when $x \to \infty$, the function $f$ is bounded above by a scalar multiple of $g$. Note that the definition says nothing about the limit of $x \to 0$. In particular, one has that
$$f(x) = x + 1 \in O(x)$$
and $f(x)$ tends to $1$ as $x \to 0$.