function asymptotic where $f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$ If $a$ and $b$ are positive real numbers, and if $f(x)$ has the following asymptotic property
$f(x) = \frac{a + O(\frac{1}{\sqrt{x}})}{b + O(\frac{1}{\sqrt{x}})}$
then is the following true?
$f(x) = \frac{a}{b} + O(\frac{1}{\sqrt{x}})$
This might look like homework but it isn't.
 A: Yes.  One way to see this is to actually do the long division (like the kind you learned in elementary school)!  Unfortunately, typesetting that in full on this forum will overtax my LaTeX powers.
Anyway, dividing $b + O\left(\frac{1}{\sqrt{x}}\right)$ into $a + O\left(\frac{1}{\sqrt{x}}\right)$ yields $\frac{a}{b}$ with a remainder of $O\left(\frac{1}{\sqrt{x}}\right)$.  So we have 
$$\frac{a + O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + \frac{O\left(\frac{1}{\sqrt{x}}\right)}{b + O\left(\frac{1}{\sqrt{x}}\right)} = \frac{a}{b} + O\left(\frac{1}{\sqrt{x}}\right),$$
since $b + O\left(\frac{1}{\sqrt{x}}\right) = O(1).$
A: It is true.  In the spirit of epsilon/delta, you are challenged to prove that $|f(x)-\frac{a}{b}| \lt \frac{M}{\sqrt{x}}$ for $x\gt x_0$ where your challenger gives M and you have to find an $x_0$ that works.  But you get to challenge back saying the numerator should be within $\frac{N}{\sqrt{x}}$ of $a$ and similarly the denominator should be within $\frac{P}{\sqrt{x}}$ of $b$.  So take $N=\frac{M}{2b}$ and $P=\frac{aM}{2b^2}$ and take the larger of the $x_0$'s that come back.
