If for some $n\in\mathbb{N}$, $\lim\limits_{x\to\infty}\frac{f(x)}{x^n}$ exists, then $f$ is rational

Question

I don't know if this statement is true. Let $F$ be a function and suppose $n>0$, $n\in\mathbb{N}$ is the greatest such that there exists $L\mathbf{\neq 0}$ such that $\lim\limits_{x\to\infty}\frac{f(x)}{x^n}=L$. Does this imply $f$ is a rational function?

Edit: I have realized that the above is not true, so I will modify it. If the above conditions hold, and $f$ is continuous, can $f$ be written as a sum of a rational function and a function $g$ such that $\frac{g(x)}{x^n}\to 0$?

Answer 1

The edited version has a trivial affirmative answer: if $f$ is rational, than you're done; if it is not, write it as $L x^n + (f- Lx^n)$. Obviously, $Lx^n$ is rational. If $g=f - L x^n$, then $\lim \limits _{x \to \infty} \frac {g(x)} {x^n} = \lim \limits _{x \to \infty} \frac {f(x)} {x^n} - L = 0$.

Answer 2

Take $f(x) = x + \exp(-x)$ and $n = 1$. The limit equals 1 but $f(x)$ isn't a rational function.