$x \mathcal{o}\left(\frac{1}{x^3}\right) =\mathcal{o}\left(\frac{1}{x^2}\right) $? Is this correct:
$$x \mathcal{o}\left(\frac{1}{x^3}\right) = \mathcal{o}\left(\frac{1}{x^2}\right)$$
or the expression $x \mathcal{o}\left(\frac{1}{x^3}\right)$ cannot be simplified?
Here $\mathcal{o}(\cdot)$ stands for little $\mathcal{o}$ of $\ \cdot$.

$f(x) = o\left(\dfrac{1}{x^2} \right)$ if $\lim_{x \rightarrow \infty} \ x^2f(x) = 0.$
Define $g(x) = \dfrac{1}{x}f(x)$, so $g(x) = \dfrac{1}{x}o\left(\dfrac{1}{x^2}\right)$, but
$$\lim_{x \rightarrow \infty} \ x^3g(x) =\lim_{x \rightarrow \infty} \ x^2f(x) = 0$$ so $g(x) = o\left(\dfrac{1}{x^3}\right)$.
