I was wondering if the order between multiplication and little o can be exchanged, for example, $$x \times o\left(\frac{y}{x}\right) = o(y)?$$
I am a little confused.
In the example, I need to consider the case when $x$ goes to infinity, and don't know if eliminating $x$ will be a good idea. As far as I know, it is correct that $\lim_{x\rightarrow \infty} x \times o\left(\frac{y}{x}\right) =0$, but I can't get $\lim_{x\rightarrow \infty} o(y) =0$. So I think $$x \times o\left(\frac{y}{x}\right) = o(y)$$ is not true?
Added: My case is that given $x \times o_{t\rightarrow \infty}(t^2)$ and $t=\sqrt{y/x}$, so I get $x \times o_{y/x \rightarrow \infty}(y/x)$. Now with this meaning and my consideration when x goes to infinity, is $x \times o_{y/x \rightarrow \infty}(y/x)$ same as $o_{y \rightarrow \infty}(y)$ or as $o_{y/x \rightarrow \infty}(y)$? Is it right that $\lim_{x\rightarrow \infty} f(x,y)=0, \forall f \in x \times o(y/x)$?
Thanks!