$f(x)\in O(\frac{1}{x})$ implies $\log(f(x))\in O(\frac{1}{x^2})$?

Question

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?

Answer 1

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 $x \to 0$.