Limit of $(x^n/e^x)$ when x tends to plus infinity I was working on this and tried to solve this using L'Hopital's law
BUT:
derivative on $x^n$ is say , $g(x)=$nx^(n-1)$ $where as n is not defined in the question,
we can't say that $g(x)$ tends to plus infinity or minus infinity .
Then I tried to expand the $e^x$ and I am stuck there , can anyone please explain me how to do it?
Thank you so much !
 A: Without L'Hospital : with  the power series of $\exp$ we get for $x>0$:
$e^x> \frac{x^{n+1}}{(n+1)!}.$
Hence
$0 < \frac{x^n}{e^x}<\frac{(n+1)!}{x}.$
A: If you apply L'Hopital's Rule $n$ times you will be left with $\lim \frac {n!} {e^{x}}$ which is $0$. 
A: Since $n$ is constant with respect to the limit, you can just "pull it out" and evaluate the limit that remains. Thus L'Hopital's gives
$$\lim_{x\to \infty} \frac{x^n}{e^x} = \lim_{x\to \infty} \frac{nx^{n-1}}{e^x} = n \lim_{x\to \infty} \frac{x^{n-1}}{e^x}$$
You still end up having to differentiate $n-1$ further times though, but afterwards the result is clear.

Of course, this is on the premise $n$ is a positive integer, which I assume given the nature of your question. This strategy doesn't quite work out so nicely otherwise and you would need to make appeals to other methods of solving this. Asymptotics would work, since the exponential grows faster than any polynomial function.
A: You can use the fact that
$$
\frac{x^n}{e^x}=\left(\frac{x}{e^{x/n}}\right)^{\!n}=
n^n\left(\frac{y}{e^y}\right)^{\!n}
$$
where $y=x/n$. It's not even necessary that $n$ is an integer, provided it is positive; then
$$
\lim_{x\to\infty}\frac{x^n}{e^x}=
\lim_{y\to\infty}n^n\left(\frac{y}{e^y}\right)^{\!n}
$$
and you're done when you have proved that
$$
\lim_{y\to\infty}\frac{y}{e^y}=0
$$
which is a simple application of l'Hôpital (or with several other methods).
