$f(x)=o(\ln^2(x)/x^2)\Rightarrow f=o(1)$? Assume that $f(x)=o\left(\frac{\ln^2(x)}{x^2}\right)$ as $x\to +\infty$. Does this imply that $f(x)=o(1)$ as $x\to +\infty$?
What I need to show is $\lim_{x\to +\infty}\lvert f(x)\rvert=0$, right?
Since $f(x)=o\left(\frac{\ln^2(x)}{x^2}\right)$ as $x\to +\infty$ means that
$$
\lim_{x\to +\infty}\left(\frac{x^2}{\ln^2(x)}\cdot \lvert f(x)\rvert\right)=0 \tag{$\ast$}
$$
but, by L'Hôspital,
$$
\frac{x^2}{\ln^2(x)}\to +\infty\text{ as }x\to +\infty
$$
and hence we need to have $\lim_{x\to +\infty}\lvert f(x)\rvert=0$ in order to have $(*)$.
Am I right?
 A: Since $f=o\left(\frac{\log^2(x)}{x^2}\right)$, then for all $\epsilon>0$, there exists a number $B$ such that 
$$|f(x)|\le \epsilon \frac{\log^2(x)}{x^2}$$  
whenever $x>B$.
And since $\log^2(x)\le x^2$ for $x>1$, then $|f(x)|\le \epsilon$ for $x>B$, which by definition means that $f=o(1)$.
A: For what it's worth: your reasoning is correct, and basically boils down to the transitivity of $o(\cdot)$ (more at the end).  A simple way to make it formal is as follows:
From $(\ast)$ and your observation that $$\lim_{x\to\infty} \frac{x^2}{\ln^2 x} = \infty \tag{$\ast\ast$}$$
you are done.
Indeed, $(\ast\ast)$ implies that for some $A\geq 0$, for any $x\geq A$ we have  $\frac{x^2}{\ln^2 x}\geq 1$. And then,
for $x\geq A$
$$
0 \leq \lvert f(x)\rvert = 1\cdot \lvert f(x)\rvert \leq \frac{x^2}{\ln^2 x}\lvert f(x)\rvert \xrightarrow[x\to\infty]{(\ast)} 0
$$
so that $f(x) = o(1)$ as desired. 

Transitivity of Landau notation:

Let $a\in[-\infty,\infty]$. If, when $x\to a$, we have $f(x) = o(g(x))$ and $g(x) = o(h(x))$, then $f(x) = o(h(x))$.

Since $f(x)=o\left( \frac{\ln^2 x}{x^2} \right)$ and $\frac{\ln^2 x}{x^2} = o(1)$, this implies your result.
