Limiting behavior of a fraction Suppose we have a fraction of the form
$$g(x)=\frac{1}{a_0 + a_1 x + \dots + a_n x^n + f(x)} \; ,$$
where $a_0 \neq 0$ and $f(x) = O(x^{n+1})$, but not necessarily polynomial or differentiable. Is there a general expression for $b_i$ such that
$$g(x)=b_0 + b_1 x + \dots + b_m x^m + O(x^{m+1})$$
holds for $m$ as high as possible?
I got as far as:
$$ g(x) = \frac{1}{a_0} - \frac{a_1}{a_0^2}x + O(x^2) \; ,$$
for $n=1$, but I do not see the general picture.
 A: I don't know if
there is a general formula,
but you can always
multiply and get an iterative form
for the successive coefficients.
In this case,
you want
(I'll ignore the $O(x^{...})$
and the extra $a_i$ and $b_j$  for now)
$\begin{align}
1
&= \left(\sum_{i=0}^n a_i x^i\right)
\left(\sum_{i=0}^m b_j x^j\right)\\
&= \sum_{i=0}^n \sum_{i=0}^m a_i b_jx^{i+j}\\
&= \sum_{k=0}^{n+m} \sum_{i=0}^k a_i b_{k-i} x^k\\
&= \sum_{k=0}^{n+m} x^k \sum_{i=0}^k a_i b_{k-i} \\
\end{align}
$
Equating the constant term,
$1 = a_0 b_0$,
so $b_0 = 1/a_0$.
For all the other terms,
$0 
= \sum_{i=0}^k a_i b_{k-i} x^k
= a_0 b_k +\sum_{i=1}^k a_i b_{k-i}
$
so
$b_k = -\frac1{a_0}\sum_{i=1}^k a_i b_{k-i}
$
You keep on doing this
until $k>n$,
when $a_k$ is no longer known.
There is absolutely nothing original here,
but I hope it helps.
A: First off: write the polynomial part of your denominator as $P(x)$. We can write
$$
\frac{1}{P(x)+f(x)}=\frac{1}{P(x)}\cdot\frac{1}{1+\frac{f(x)}{P(x)}},
$$
provided $P(x)\neq 0$ (which is true for $x$ sufficiently small, since $P(0)=1$); but since $P(x)\rightarrow1$ as $x\rightarrow0$ and $f(x)=O(\lvert x\rvert^{n+1})$, it is true that $\frac{f(x)}{P(x)}=O(\lvert x\rvert^{n+1})$ as $x\rightarrow0$ as well. But we know that $\frac{1}{1-w}=1+O(\lvert w\rvert)$ as $w\rightarrow0$, so that
$$\tag{1}
\frac{1}{1+\frac{f(x)}{P(x)}}=\frac{1}{1-\left(-\frac{f(x)}{P(x)}\right)}=1+O\left(\left\lvert\frac{f(x)}{P(x)}\right\rvert\right)=1+O(\lvert x\rvert^{n+1})\text{ as }x\rightarrow0.
$$
So, at the end of the day, you are really only asking to find an expansion for $\frac{1}{P(x)}$.
To that end: note that as long as $a_1x+a_2x^2+\cdots+a_nx^n\in(-1,1)$, we have
$$
\begin{align*}
\frac{1}{P(x)}&=\frac{1}{1-(-1)(a_1x+a_2x^2+\cdots+a_nx^n)}\\
&=\sum_{k=0}^{\infty}(-1)^k\left(a_1x+a_2x^2+\cdots+a_nx^n\right)^k\\
\end{align*}
$$
You can write explicit formulas from here by using multi-indices or things of that nature; at the end of the day, though, it is a mess!
It is not bad for fixed, specific $n$ though.  For instance: when $n=1$, you get
$$
\frac{1}{1+a_1x}=\sum_{k=0}^{\infty}(-1)^k(a_1x)^k=1-a_1x+a_2x^2+\cdots
$$
Since we only know that $f(x)=O(\lvert x\rvert^2)$, you can stop this at $1-a_1x+O(\lvert x\rvert ^2)$. 
When $n=2$, we have
$$
\frac{1}{1+a_1x+a_2x^2}=\sum_{k=0}^{\infty}(-1)^k(a_1x+a_2x^2)^k.
$$
You only need to consider the first and second order terms:
$$
\frac{1}{1+a_1x+a_2x^2}=1-(a_1x+a_2x^2)+(a_1x+a_2x^2)^2+O(\lvert x\rvert^3),
$$
and you can expand it out from there. This works because the $k=3$ term of the summand only involves powers of $x$ that are $3$ or greater.
From here, it is a matter of handling the general $a_0\neq 0$ case; but, you can do this by factoring out $a_0$, yielding a polynomial with constant term $1$, and then applying the previous.
