What is $\lim_{x\to\infty} x^ne^{−x^2}?$ What is $$\lim_{x\to\infty} x^ne^{−x^2}?$$ for any $n$ I think its $0$ because the exponential will go faster to $0$ than the nth power go to +∞, but I don't see which theorem to apply to reach the result.
Thank you!
 A: Consider also this amusing way to see it:
$$\lim_{x\to +\infty} x^n e^{-x^2} = \lim_{x\to +\infty} \dfrac{x^n}{e^{x^2}}$$
We can use De L'Hôpital rule here. Clearly using it once won't give you any immediate result (if we don't take into account hierarchy orders of rapidity and so on). Indeed after using H once:
$$\lim_{x\to +\infty} \dfrac{x^n}{e^{x^2}} \longrightarrow \lim_{x\to +\infty} \dfrac{n x^{n-1}}{2x e^{x^2}}$$
The interesting part is mentally use it $n$ times, where you can easily perceive by intuition that
$$\dfrac{\text{d}^n}{\text{d}x^n} x^n = n!$$
as well as
$$\dfrac{\text{d}^n}{\text{d}x^n} e^{x^2} = c(n) e^{x^2} P^n(x)$$
where $P^n(x)$ represents a polynomial of degree $n$ and $c(n)$ is a constant depending on $n$ only.
Whence:
$$\lim_{x\to +\infty} x^n e^{-x^2} = \lim_{x\to +\infty} \dfrac{n!}{c(n) e^{x^2} P^n(x)} \to 0$$
A: We have $${x^n\over e^{x^2}}=\left ( {x\over e^{x^2/n}}\right )^n$$ Therefore it suffices to show that $$\lim_{x\to \infty} {x\over e^{x^2/n}}=0$$ We can use $e^t\ge t,$  for $t\ge 0$, i.e. $e^{x^2/n}\ge x^2/n.$ Hence $$0\le {x\over e^{x^2/n}}\le {n\over x}$$
