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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?

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    Does $o$ stand for "little o"? – 2017-01-27
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    Yes, the Little o Landau Symbol. – 2017-01-27

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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 $f=o(1)$.

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    Your answer is very nice, thank you. Is *my* argumentation okay or is it unsatisfying? – 2017-01-27
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    You're welcome. My pleasure. Your argument works. – 2017-01-27
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    The first line should probably read "then, for all $\epsilon > 0$ there exists $A\geq 0$ such that, for all $x\geq A$" instead. – 2017-01-27
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    @clementc. Yes, of course. I was hastily writing this and dropped that. I've edited. Thank you and Happy New Year! -Mark – 2017-01-27
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    Happy new year (belated)! – 2017-01-27
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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.