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 !
 A: If you do not want to use L'Hopital's rule, this is another posibility. If $x>0$ then
$$
e^x=\sum_{k=0}^\infty\frac{x^k}{k!}>\frac{x^{n+1}}{(n+1)!}\implies e^{-x}<\frac{(n+1)!}{x^{n+1}}.
$$
A: Theo Bendit gave the hint. Write $$A=\frac{x^n}{e^x}=\frac uv$$ Apply L'Hospital a first time $$\frac {u'}{v'}=\frac{nx^{n-1}}{e^x}$$ Repeat $$\frac {u''}{v''}=\frac{n(n-1)x^{n-2}}{e^x}$$ $$\frac {u'''}{v'''}=\frac{n(n-1)(n-2)x^{n-3}}{e^x}$$After $n$ steps, you have a constant in the numerator (which you even do not need to know that it is $n!$) and still the beautiful $e^x$ in the denominator.
A: "My guess is that ... because the exponential will go faster to $0$ than the nth power go to $\infty$". This guess is actually based on the fact that $\lim_{x \to \infty}x^{n}e^{-x} = 0$ for all $n > 0$ and not the other way round.
As easy way out is to put $e^{x} = t$ so that $x = \log t$ and then $x^{n}/e^{x} = (\log t)^{n}/t$. Further if $t > 1$ then $$\log t = 2n\log t^{1/2n} \leq 2n(t^{1/2n} - 1) < 2nt^{1/2n}$$ so that $$0 < \frac{\log t}{t^{1/n}}< \frac{2n}{t^{1/2n}}$$ Using squeeze theorem we get that $(\log t)/t^{1/n} \to 0$ as $t \to \infty$ and hence $$(\log t)^{n}/t = ((\log t)/t^{1/n})^{n} \to 0$$
A: 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 $c>0,R>0$ with $|P(x)|\leq c|x|^n$ for alle $|x|\geq R$. From the series representation we directly obtain $$\exp(x)>\frac{1}{(n+1)!}x^{n+1}$$ 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.
A: One way is to use l'Hospital (see comment by @theo-bendit ), without any "calculation": $\exp{x}=\sum_{k=0}^\infty x^k/k! > \sum_{k=0}^{n+1} x^k/k!$ and then $x^n/\exp{x} < x^n / \sum_{k=0}^{n+1}  x^k/k!$ . We divide both parts of the fraction by $x^n$ and see that the limit is basically the same as that of $1/x$ for large $x$.
A: $x^n\cdot e^{-x} = e^{-x+n\ln x}$. We show: $-x + n\ln x \to -\infty$ when $x \to +\infty$. First choose $x > 3n$, then: $-x+n\ln x = -1+\displaystyle \int_{1}^x \left(\dfrac{n}{t} - 1\right)dt=-1+\displaystyle \int_{1}^{3n} \left(\dfrac{n}{t}-1\right)dt+\displaystyle \int_{3n}^x \left(\dfrac{n}{t} - 1\right)dt < C(n) -\dfrac{2}{3}\left(x-3n\right) \to -\infty$ as $x \to +\infty$. Thus: $-x+n\ln x \to -\infty \Rightarrow x^\cdot e^{-x} \to 0$. Here $C(n)$ is a constant.
A: The answer is $n!$. Bring $e^{-x}$ in denominator and differentiate $n$ times
