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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!

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1 Answer

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You have every right to be confused; little-o notation itself is problematic, and this question highlights at least two of its problems.

Let's be completely precise. We say that $f(x) \in o_{x \to \infty}(g(x))$ (note the subscript) if $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0$. This is the most precise form of little-o notation: I have not used an equals sign (because $o(g(x))$ is not a function), I have specified what variable the notation refers to, and I have even specified what its limit is. (Most people who use little-o or related notations don't bother to do any of these things, and the result can be quite confusing if you aren't aware of the context.)

Thus $f(y) \in o_{y \to \infty}(y)$ if $\lim_{y \to \infty} \frac{f(y)}{y} = 0$. In this situation, we might want to interpret

$$f(x, y) \in x o_{x \to \infty} \left( \frac{y}{x} \right)$$

to mean that

$$\frac{f(x, y)}{x} \in o_{x \to \infty} \left( \frac{y}{x} \right)$$

where $y$ is possibly a function of $x$. This is equivalent, of course, to $\lim_{x \to \infty} \frac{f(x, y)}{y} = 0$, so it is equivalent to the statement that

$$f(x, y) \in o_{x \to \infty}(y).$$

But your question is a little confusing, because you have two variables $x$ and $y$ and you didn't specify which one of them the little-o notation refers to, and you also didn't refer to $f$. Both of these are weaknesses of little-o notation itself.

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