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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.

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    $\begingroup$ e^(x^2) increases much faster than x! $\endgroup$ – user600016 Jun 11 at 1:25
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    $\begingroup$ Have you tried the same trick but replacing $e$ in the denominator with $e^x$. Because product of $x$ of $e^x$ is $e^{x^2}$. $\endgroup$ – kingW3 Jun 11 at 1:27
  • $\begingroup$ Another option is to use Stirling's Approximation for the $x!$ part $\endgroup$ – amcalde Jun 11 at 1:33
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$y = x^x$ 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 $y = e^{x^2}$?

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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 $0$.

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  • $\begingroup$ It's just $e^{x^2}$ where did you get so many e's from? $\endgroup$ – Archis Welankar Jun 11 at 12:44
  • $\begingroup$ @ArchisWelankar: come on, $e^{x^2}=(e^x)^x$. $\endgroup$ – Yves Daoust Jun 11 at 12:50
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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 $$ \lim_{x \to \infty} \left(x! e^{-x^2}\right) = 0. $$

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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 $$\frac{n!}{e^{n^2}} \to 0$$

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