# 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.

• e^(x^2) increases much faster than x! Jun 11 '19 at 1:25
• Have you tried the same trick but replacing $e$ in the denominator with $e^x$. Because product of $x$ of $e^x$ is $e^{x^2}$. Jun 11 '19 at 1:27
• Another option is to use Stirling's Approximation for the $x!$ part Jun 11 '19 at 1:33

$$y = x^x$$ 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 $$y = e^{x^2}$$?

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 $$0$$.

• It's just $e^{x^2}$ where did you get so many e's from? Jun 11 '19 at 12:44
• @ArchisWelankar: come on, $e^{x^2}=(e^x)^x$.
– user65203
Jun 11 '19 at 12:50

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 $$\lim_{x \to \infty} \left(x! e^{-x^2}\right) = 0.$$

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