How do you evaluate $\lim_{x\to\infty}(x!*e^{-x^2})$ How do you evaluate $\lim_{x\to\infty} (x!*e^{-x^2})$?
I know that $\lim_{x\to\infty} (x!*e^{-x}) = \infty$ because for large values of $x$, $\frac{1*2*3...*(x-1)*x}{e*e*e*...*e*e}= \frac{\text{A lot of e's}}{\text{less e's}}=\infty$.  In this case, everything in the numerator above $e$ contributes more than an $e$, while everything in the denominator contributes exactly one $e$.
Is there a way of using the same type of reasoning for $\lim_{x\to\infty} (x!*e^{-x^2})$?
I don't know how to approach this because there are different amounts of numbers in the numerator and denominator.
 A: $y = x^x$ certainly grows faster than $y = x!$, because
\begin{align}
x^x &= x \cdot x \cdot x \cdots x \\
&\ge x \cdot (x - 1) \cdot (x - 2) \cdots 2 \cdot 1 \\
&= x!.
\end{align}
But $y=e^{x^2}$ grows faster than $y=x^x=e^{x\ln{x}}$ because $x^2$ grows faster than $x \ln x$.
Knowing this, what can we conclude when we compare $y = x!$ to the reciprocal of $y = e^{x^2}$?
A: We have that
$$\frac{x!}{e^{x^2}}=\frac{1\cdot2\cdot3\cdots(x-1)\cdot x}{e^x\cdot e^x\cdot e^x \cdots e^x*e^x}=\frac{1}{e^x}\cdot\frac{2}{e^x}\cdots \frac{x}{e^x}$$
And each of term goes $\to 0$ as $x\to \infty$ hence the limit is $0$.
A: Stirling's approximation says that
$$
\ln \left[ x! e^{-x^2} \right] = \ln x! - x^2 \approx (x \ln x - x) - x^2
$$
and since this quantity goes to $-\infty$ as $x \to \infty$, we conclude that
$$
\lim_{x \to \infty} \left(x! e^{-x^2}\right) = 0.
$$
A: Another approach: we have that
$$\sum_{n \in \mathbb{N}} \frac{n!}{e^{n^2}}$$
converges by the ratio test:
$$\frac{(n+1)!}{e^{(n+1)^2}} \frac{e^{n^2}}{n!}=\frac{n+1}{e^{2n+1}} \to 0$$
Which means that
$$\frac{n!}{e^{n^2}} \to 0$$
