Asymptotics of coefficients This is a question that asks the reader for a $strategy$ to solve a particular problem. I cannot solve this problem myself so I am looking around for general methods one might use to confront it with. Suppose
$$f(x)=a_0+a_1x+...,  g(x)=b_0+b_1x_...$$
and given
$$\lim\limits_{x\to 1^-}\frac{f^{(n)}(x)}{g^{(n)}(x)}=1$$
With the additional
$$\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$$ and $$\lim\limits_{n\to\infty}\frac{b_{n+1}}{b_n}=1 $$
for all natural $n$. Prove
$$\lim\limits_{n\to \infty}\frac{a_n}{b_n}=1$$
I would like to know your general thoughts about approaching this type of problem. Any insights however small will be much appreciated. Although this is seemingly a problem in real analysis, I have tagged complex analysis because solutions may well involve it.
 A: Under the given assumptions,
$$\lim_{n\to\infty} \frac{a_n}{b_n} = 1$$
need not hold. Counterexample:
$$\begin{align}
f(x) &= \frac{1}{1-x^2} = \sum_{k=0}^\infty x^{2k},\\
g(x) &= f(x) + e^x.
\end{align}$$
We have $\dfrac{a_n}{b_n} = 0$ for all odd $n$, but nevertheless
$$\lim_{x\to 1} \frac{f^{(n)}(x)}{g^{(n)}(x)} = \lim_{x\to 1} \frac{f^{(n)}(x)}{f^{(n)}(x)+e^x} = 1$$
for all $n$, since $\lim\limits_{x\to 1} \lvert f^{(n)}(x)\rvert = +\infty$, while $e^x$ remains bounded.
You need some further conditions to reach the conclusion.
A: Firstly suppose that $f(x)$ and $g(x)$ are two polynomials, and let their degree be $c$ and $d$, respectively. Suppose $ d \gt c$, you get
$\lim\limits_{x\to 1^-}\frac{f^d(x)}{g^d(x)} = \frac{0}{d!\cdot b_d} = 0 \ne 1$, so it's impossible that $ d \gt c$; obviously you can use the same proof to prove that it's impossible that $ c\gt d$.
Now, since $c=d$, $\lim\limits_{x\to 1^-}\frac{f^c(x)}{g^c(x)} = \frac{c!\cdot a_c}{c!\cdot b_c} = \frac{a_c}{b_c} = 1$ , so $a_c$ = $b_c$.
Similarly, $\lim\limits_{x\to 1^-}\frac{f^{c-1}(x)}{g^{c-1}(x)} = \frac{(c-1)!\cdot a_{c-1}+c!\cdot x\cdot a_c}{(c-1)!\cdot b_{c-1}+c!\cdot x\cdot b_c} = \frac{a_{c-1}+a_c}{b_{c-1}+b_c} = 1$, which means that 
$a_{c-1}+a_c = b_{c-1}+b_c$
$a_{c-1} = b_{c-1}$.
Clearly you go on with the same "strategy", and you'll obtain that $a_i$ = $b_i$ for all $i$ such that  $0\le i \le c$, so $\lim\limits_{n\to \infty}\frac{a_n}{b_n}=1$.
