Confused about dealing with Big $\mathcal{O}$ in functions For the following defined functions 
$$f(x) = \frac{1}{x}+\frac{n-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)$$ 
and 
$$y(x) = c\left(\frac{1}{x}+\frac{m-1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)$$ 
where $m > n > 0$ and $c$ is large.
I want to bound $f(x)-y(x))$.
I thought that $f(x)-y(x) \leq \frac{1}{x}+\frac{n-1}{x^2}- c\left(\frac{1}{x}+\frac{m-1}{x^2} \right)$, since we are talking about an approximation!
Is this true?
After that, I found this question where the answer states that 

For every $x\geq0$, $f(x)-y(x) \leq  f(x)$

Is this the tighter bound? if so, how can I get rid of the $\mathcal{O}$ in $f(x)$?
 A: Regarding your first question

I thought that $ f(x) − y(x) \leq \frac{1}{x} + \frac{n−1}{x^2} −c (\frac{1}{x} + \frac{m−1}{x^2}) $
  , since we are talking about an approximation!
Is this true?

No. Consider for example $ f(x) = \frac{1}{x} + \frac{n-1}{x^2} + \frac{c+1}{x^3}, g(x) = c(\frac{1}{x} + \frac{m-1}{x^2} + \frac{1}{x^3}) $. Then you get $ f(x) - g(x) =  \frac{1}{x} + \frac{n-1}{x^2} - c(\frac{1}{x} + \frac{m-1}{x^2}) + \frac{1}{x^3} > \frac{1}{x} + \frac{n-1}{x^2} - c(\frac{1}{x} + \frac{m-1}{x^2}) $ 
A: You don't get rid of the $\mathcal{O}$, but rather the argument is, that for some reason you do not need to keep close track of it. I feel like there should be some kind of rules how to calculate with them properly. They could look like this:
$\mathcal{O}(f(x)) = c\mathcal{O}(f(x))$
$\mathcal{O}(f(x)) \pm \mathcal{O}(f(x))   = \mathcal{O}(f(x))$
and that would lead to
$f(x)-y(x) \leq \frac{1}{x}+\frac{n-1}{x^2}- c\left(\frac{1}{x}+\frac{m-1}{x^2} \right) + \mathcal{O}\left(\frac{1}{x^3}\right)$
