I am considering a smooth-enough real-valued function $ f: (0,1) \to (0,\infty) $ such that


1. $ f $ is decreasing,
2. $\lim\_{x\rightarrow0^{+}}f(x)=\infty $,
3. $ x \mapsto x^{2} f'(x) $ is decreasing,
4. $\lim\_{x\rightarrow0^{+}}x^{2} f'(x)=0 $.



> 
> **QUESTION.** Under these constraints, does the limit
>  $$ \lim\_{x\rightarrow0^{+}}\dfrac{\log(f(x))}{\log(x)}$$ 
>  exist?
> 
> 
> 


Thanks a lot to anybody who has any thoughts or counterexamples or spent time reading this question!



On the existence of $ \lim_{x \to 0^{+}} \frac{\log(f(x))}{\log(x)} $ under some constraints
Here's one way to get a counterexample.


Let $f(x) = g(1/x)$, so $g'(t) = -(1/t^2) f'(1/t)$.
Your conditions say that as $t \to +\infty$, we have $g$ positive and increasing to $\infty$, with $g'$ decreasing to $0$.
We can choose sequences $t\_n$ and $y\_n$ such that 


1. $y\_n = t\_n^{1/4}$ if $n$ is even, $t\_n^{1/2}$ if $n$ is odd.
2. The slopes $s\_n = (y\_{n+1}-y\_n)/(t\_{n+1}-t\_n)$ are positive and decreasing to $0$ with $n$
3. $t\_n \to \infty$ as $n \to \infty$.


Take the piecewise linear function whose graph has vertices $(t\_n, y\_n)$; after smoothing off the corners we get a suitable function $g$ where 
$\log(g(t))/\log(t) = -\log(f(1/t))/\log(1/t)$ oscillates infinitely often between $1/4$ and $1/2$. 


