Let $P(x)$, a polynomial which isn't the zero-polynomial.
I want to prove the following limits
$$\lim \limits_{x\to\infty} \left|P(x)\right|e^{-x} = 0$$ $$\lim \limits_{x\to-\infty} \left|P(x)\right|e^{-x} = \infty$$
Now, I already know that exponential is growing faster than polynomial but how to show it?
Suppose $P(x) = \sum_{i=0}^n a_ix^i$ for some $a_0,\ldots,a_n$. Then, $|P(x)| \leq \sum_{i=0}^n |a_i||x|^i$. Suppose $x \geq 1$. Then, $|x|^i \leq x^n$ for every $0\leq i \leq n$. Hence, $|P(x)| \leq C x^n$ whenever $x \geq 1$, where $C = \sum_{i=0}^n |a_i|$. Now, $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!} \geq \frac{x^{n+1}}{(n+1)!}$ so that $e^{-x} \leq (n+1)!x^{-(n+1)}$. Combining with the bound above, $|P(x)|e^{-x} \leq C(n+1)! \frac{1}{x}$ when $x \geq 1$. This leads to the first limit. The second can be done using similar arguments.
You can just use the Bernoulli inequality, for which $$a_n = \left(1+\frac{x}{n}\right)^n $$ is an increasing sequence that converges to $e^x$. Assuming that $$ P(x) = \sum_{j=0}^{m} a_j x^j, $$ take $M=\sum_{j=0}^m |a_j|$ and $n=m+1$. Then, assuming $x>1$: $$\frac{|P(x)|}{e^x}\leq \frac{M x^{n-1}}{\left(1+\frac{x}{n}\right)^n}=\frac{M}{x}\left(\frac{nx}{n+x}\right)^n\leq \frac{M n^n}{x},$$ hence the limit is zero by squeezing ($Mn^n$ can be huge but it is constant). On the other hand, if $x$ tends to $-\infty$ both $|P(x)|$ and $e^{-x}$ increases without bound given that $|x|$ is big enough, hence: $$\lim_{x\to -\infty}|P(x)|\,e^{-x} = +\infty.$$
First prove the base case of $P(x)$ nonzero constant. Now for inductive step we assume $P(x)$ is a not a constant, hence $P^{\prime}(x)$ is a nonzero polynomial. By induction hypothesis any polynomial of lower degree already satisfy $\lim\limits_{x\rightarrow\infty}\frac{P(x)}{e^{x}}=0$ and $\lim\limits_{x\rightarrow-\infty}\frac{P(x)}{e^{x}}=\pm\infty$ (these inequality are stronger than the one in the question, hence more suitable for induction; once you prove them, adding in the absolute value is just 1 line)
Since $P^{\prime}(x)$ have finite root its sign will eventually be fixed so $P(x)$ is eventually monotone so $P(x)$ either diverge to $\pm\infty$ or to a finite value as $x\rightarrow\pm\infty$.
If $\lim\limits_{x\rightarrow\infty}P(x)\not=\pm\infty$ then we immediately have $\lim\limits_{x\rightarrow\infty}\frac{P(x)}{e^{x}}=\frac{\lim\limits_{x\rightarrow\infty}P(x)}{\lim\limits_{x\rightarrow\infty}e^{x}}=0$. Otherwise $\lim\limits_{x\rightarrow\infty}\frac{P(x)}{e^{x}}=\lim\limits_{x\rightarrow\infty}\frac{P^{\prime}(x)}{e^{x}}$ by L'Hospital rule, and since $P^{\prime}(x)$ have lower degree, this complete the induction step.
If $\lim\limits_{x\rightarrow-\infty}P(x)\not=0$ then we immediately have $\lim\limits_{x\rightarrow-\infty}\frac{P(x)}{e^{x}}=\frac{\lim\limits_{x\rightarrow-\infty}P(x)}{\lim\limits_{x\rightarrow-\infty}e^{x}}=\pm\infty$. Otherwise, $\lim\limits_{x\rightarrow-\infty}\frac{P(x)}{e^{x}}=\lim\limits_{x\rightarrow-\infty}\frac{P^{\prime}(x)}{e^{x}}$ by L'Hospital, and since $P^{\prime}(x)$ have lower degree, this complete the induction step.
Off the top of my head:
$$|P(x)|e^{-x}\to0~~\Leftarrow~~ x^ne^{-x}\to0~~\Leftarrow~~xe^{-x}\to0~~\Leftarrow~~\log x-x\to-\infty$$
which happens if $\int_1^x(\frac{1}{u}-1)du\le \int_1^2(\frac{1}{u}-1)du+\int_2^x(\frac{1}{2}-1)du$ diverges, which should be clear.
The second limit to show should be pretty obvious, since $|P(x)|$ is bounded below if sufficiently far away from its zeros and the term $e^{-x}$ tends to $\infty$ as $x\to-\infty$.
One could also consider l'Hospital but that assumes the limit exists.
