Factorial in limit

Question

Evaluate the limit of $\frac{x^x}{x!}$ as x approaches infinity. In this problem,how to think of $x$? The solution I came across use $x!=x(x-1)(x-2)\cdots1$ to show the inequality that limit $L \le \frac{1}{x}$. But the limits we are familiar with readily assume that $x$ is a real number. If that's the case,how does the solution assuming the elementary definition of factorial work where irrational numbers dont follow this definition?

Answer 1

The same approach works. It's just that we can only do it to the integer part. Split your real number $x=n+s$ where $n\in\mathbb{N}$. Then $$\frac{\Gamma(x+1)}{x^x}=\frac{\Gamma(1+s)(1+s)(2+s)\dots(n+s)}{x^nx^s}$$ $$=\frac{\Gamma(1+s)}{x^s}\frac{(1+s)(2+s)\dots(n+s)}{x^n}$$ The second part satisfies the same inequality you got before, so you get that $$f(x)\le\frac{\Gamma(1+s)}{x^s}\frac{1+s}{x}$$ $\Gamma$ is continuous on $[1,2]$, so it has a maximum, which is strictly positive, so you can replace it with a constant $c$. And $1+s\le 2$. Then you have $$f(x)\le\frac{2c}{x^{1+s}}$$ Which clearly goes to $0$.