$\dfrac{O(1)}{x} = O(\dfrac{1}{x})$?

Question

I just saw a lecture where the professor did this simplification in a proof but did not comment it further. How would you go about provning the statement in the title?

Answer 1

For the $x\to\infty$ case, the following statements are equivalent:

• $f\in\frac{O(1)}{x}$
• $xf\in O(1)$
• There exist $x_0,\,M$ with $M>0$ such that for all $x>x_0$, $|xf(x)|<M$
• There exist $x_0,\,M$ with $M>0$ such that for all $x>x_0$, $|f(x)|<M|1/x|$
• $f\in O(1/x)$

You can handle $x\to0$ or $x\to0^+$ similarly.

Answer 2

Can this be considered a proof: let $h(x) = \dfrac{O(1)}{x} \Leftrightarrow x\cdot h(x) = O(1)$ meaning specifically $x\cdot h(x) \le K\cdot 1\Leftrightarrow h(x) \le K\cdot \dfrac{1}{x}$ for some constant $K$. Thus $h(x) = O(1/x)$.