Restrictions imposed on function by limit condition If a continuous function $f(A)$ satisfies the following conditions:


*

*$$\lim_{A\to \infty}e^{-A}f(A)=0$$

*$$\lim_{A\to \infty}f(A)=\infty$$
Then, what are the restrictions on $f(A)$?
I suspect the restrictions are related to something about the speed of growth, like maybe $f(A) \le O(N^2)$. But I really lack knowledge about stuff like this, and after searching in the internet for a few days I still got no useful results.
Thanks for any help in advance.
 A: Having the limit$\lim\limits_{A\rightarrow\infty}e^{-A}f(A)=0$ means that for every $\varepsilon>0$ there exists $X$ such that $x\geq X$ implies $e^{-A}f(A)<\varepsilon$, that is $f(A)<\varepsilon\, e^{A}$. So your two conditions mean that $f(A)$ grows to infinity, but can be eventually dominated by an exponential for sufficiently large $A$.
In particular, the exponential has a Taylor series:
$$e^A=\sum_{n=0}^{\infty} \frac{A^n}{n!},$$
so any power function $f(A)=A^\gamma$ ($\gamma$ constant) is a solution since it will be eventually dominated by $e^A$ because of the higher powers. For instance, any polynomial $p(A)=\sum_{n=0}^{N} a_nA^n$ works.
However note that powers and polynomials are not the only solutions, one can find many other examples. For instance, since you only require $f$ to be continuous, functions that are piece-wise polynomial work as well. More generally, given a solution $f$, you can get another solution $g$ by altering the original function continuously on some interval which is bounded above, since we only care about what the function does "at infinity". Ultimately, any linear combination of solutions will also be a solution.
