What is the limit $x\to\infty$ of $x^n \cdot e^{-x}$? What is the limit $x\to+\infty$ of $x^n \cdot e^{-x}$ ?  My guess is that it is $0$ because the exponential will go faster to $0$ than the $n$th power go to $+\infty$, but I don't see which theorem to apply to reach the result.

Thanks for any help !
This can be shown without using L'Hospital as follows:

Let $$\exp:\mathbb R\rightarrow\mathbb R,~x\mapsto\sum\limits_{n=0}^\infty \frac{1}{n!}x^n$$ and $P(x)$ a polynomial. We show that $$\lim\limits_{x\to\infty}\frac{P(x)}{\exp(x)}=0.$$

Let $n$ be the degree of $P$, then there exists $c&gt;0,R&gt;0$ with $|P(x)|\leq c|x|^n$ for alle $|x|\geq R$. From the series representation we directly obtain $$\exp(x)&gt;\frac{1}{(n+1)!}x^{n+1}$$ for all $x\geq 0$. Using this we get $$\left|\frac{P(x)}{\exp(x)}\right|\leq \frac{cx^n}{\frac{1}{(n+1)!}x^{n+1}}=\frac{c(n+1)!}{x}$$ for all $x\geq R$. As $$\lim\limits_{x\to\infty}\frac{c(n+1)!}{x} =0$$ we are done.
The answer is $n!$. Bring $e^{-x}$ in denominator and differentiate $n$ times