$f(x)\in O(\frac{1}{x})$ implies $\log(f(x))\in O(\frac{1}{x^2})$? Consider a function $f(x):(0, \infty) \rightarrow \mathbb{R}$. 
Suppose $f(x)\in O(\frac{1}{x})$ as $x\rightarrow 0$ where Big O notation is described here.
Is it true that
$$
\log(f(x))\in O(\frac{1}{x^2})
$$
as $x \rightarrow 0$
If Yes, could you help me to show it?

It is not true. For example, consider
$$
f(x) = \exp(-1/x^4).
$$
Then $f(x) \to 0$ as $x \to 0$, so certainly $f(x) \in O(1/x)$ as $x \to 0$. But definitely
$$
\log f(x) \notin O(1/x^2)
$$
as $x \to 0$.
