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.
 A: 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 $|x| \geq \max{\{1,\frac{|a_{n-1}| + \cdots + |a_{0}|}{|a_{n}|}\}}$.
A: Write $n=e^x$ to conclude that it suffices to show that $e^x > x^2$ for all $x$ sufficiently large. Now, take $\log$s...
A: Write $P(x) = \displaystyle\sum_{n=0}^k a_n x^n$, and note $|P(x)| \leq k|a_k|x^k$ for all $x \geq 1$.
Hint: 
$$\begin{align}
|P(x)| \leq C_k x^k &\Longrightarrow \log P(x) \leq \log (C_k x^k)
\\
&\Longrightarrow \log P(x) \leq C_k + k \log x
\\
&\Longrightarrow \log P(x) \leq x
\\
&\Longrightarrow P(x) \leq e^{x}
\end{align}$$
A: Since limit of $P(n)/n^k$ is finite (k being the degree), it suffices to show that $e^n/n^k$ tends to infinity.
Since $\lim_{n\to\infty} \frac{(n+1)^k}{n^k}=1$, there exists $n_0$ such that $n>n_0$ implies $\frac{(n+1)^k}{n^k}\le 2$. 
From this you can easily see (or show by induction) that
$\frac{e^n}{n^k} \ge \frac{e^{n_0}}{n_0^k}  \left(\frac2e\right)^{n-n_0}$
whenever $n\ge n_0$. (Just notice the quotient of two consecutive terms.)
Hence  $\frac{e^n}{n^k}=C_k \left(\frac2e\right)^n$, where $C_k$ is a constant depending on $n_0$ (and thus on $k$) but not on $n$. This is what I had in mind when I posted the above comment. And it's pretty intuitive too - you simply compare the $n^k$ to the geometric progression by comparing the quotients
