calculus- limit includes polynomial Good morning,
I would like your help with proving that
$$\lim_{n \to \infty} \frac{e^n}{P(n)} = \infty,$$
where $P(n)$ is a polynomial of degree at least $1$.
Thank you very much.

Assume that $p(x) = a_{n}x^n + \cdots + a_1 x + a_0$ with $a_{n} \neq 0$ and $n \geq 1$.
Claim. For $|x|$ large enough we have
$|p(x)| \leq 2|a_{n}||x|^n.$
On the other hand we see that $e^x \geq \frac{x^{n+1}}{(n+1)!}$ for $x \geq 0$ from the defining series of $e^x$. Using the claim thus gives
$$ \left\vert \frac{e^x}{p(x)}\right\vert \geq \frac{|x|}{2|a_{n}|(n+1)!}$$
from which $\left\vert \frac{e^x}{p(x)}\right\vert \; \xrightarrow{x \to \infty} \; \infty$ follows immediately (in writing up your own solution be careful with the sign of $a_{n}$)!
To see the inequality $|p(x)| \leq 2|a_{n}||x|^n$ note that for $|x| \geq 1$ we have $|x|^{n-1} \geq |x|^k$ for $k = 0, \ldots, n-1$ so that
$$
|p(x)| \leq |a_{n}||x|^n + |a_{n-1}||x|^{n-1} + \cdots + |a_1||x| + |a_{0}| \leq 
|a_{n}||x|^n + (|a_{n-1}| + \cdots + |a_{0}|)|x|^{n-1}
$$
and it is now easy to check that the claimed inequality holds for $|x| \geq \max{\{1,\frac{|a_{n-1}| + \cdots + |a_{0}|}{|a_{n}|}\}}$.
