a question on $O$- notation I'm trying to understand $O$-notationbetter. I've found a really helpful answer : Big O notation, $1/(1-x)$ series
But I have trouble with quotient. Let me discuss about this example function:
$$f(x)=a+bx+\frac{c+dx}{e+fx^2}e^{ax}$$
If I calculate for the zero order error $O(x^0)$, I'll get
$$a+\frac{c}{e}=0$$
...for the first order error $O(x^1)$, I'll get 
$$b+\frac{d}{2fx}ae^{ax}|_{x=0}=0$$
which seems a bit or more then a bit silly:(
Could you please explain me how to calculate for the first order error $O(x^1)$?
 A: The "zero-order" approximation of $f$ near $x=0$ is given by $f(0) = a+\frac{c}{e}$. Therefore, it is often written that near 0, 
$$f(x) = a + \frac{c}{e} + O(x).$$
To get a first-order approximation, we expand $f$ by its Taylor series out to the linear term:
$$f(x) = a+\frac{c}{e} + xf'(0) + O(x^2) = a+\frac{c}{e}+\frac{ac+be+d}{e}x + O(x^2).$$
And so on for higher-order approximations of $f$.
Some items to note, if you want to make the above formal:
1) We're assuming $f$ is analytic near $0$.
2) Here $O(x)$ is referring to asymptotic behavior as $x\to 0$, instead of the usual $x\to\infty$.
A: Taylor series get you a long way in practice. Using the Taylor series for the exponential function and $\frac{1}{1 + x}$ you can get as many order terms exactly as you need
$$\begin{align} f(x) &=a+bx+\frac{c+dx}{e+fx^2}e^{ax} \\
&= a + bx + \frac{1}{e} \cdot \frac{c+dx}{1 + \frac{f}{e}x^2} \cdot \left(1 + ax + \frac{a^2}{2} x^2 + O(x^3)\right) \\ 
&= a + bx + \frac{1}{e} \cdot (c+dx) \cdot \left(1 - \frac{f}{e}x^2 + O(x^4)\right) \cdot \left(1 + ax + \frac{a^2}{2} x^2 + O(x^3)\right) \\ 
&= a + bx + \frac{1}{e} \cdot \left(c + (ac + d)x + \left(\frac{a^2 c}{2} + ad - \frac{cf}{e}\right)x^2 + O(x^3)\right) \\
&= \left(a + \frac{c}{e}\right) + \left(b + \frac{ac + d}{e}\right)x + \left(\frac{a^2 c}{2e} + \frac{ad}{e} - \frac{cf}{e^2}\right)x^2 + O(x^3). \end{align}$$
