What value of c makes this true?

Question

Since $\lim_{x \rightarrow \infty}\frac{(x)!}{x^{x}} = 0$

and

$\lim_{x \rightarrow \infty}\frac{(2x)!}{x^{x}} = \infty$

Is there a value c (or range of values) where

$\lim_{x \rightarrow \infty}\frac{(cx)!}{x^{x}} = 1$ ?

Answer 1

Using Stirling, $$\frac{(cx)!}{x^x}\approx \frac{(cx)^{cx}e^{-cx}\sqrt{2\pi cx}}{x^x}=\left(\frac{c^cx^{c-1}}{e^c}\right)^{x}\sqrt{2\pi cx}$$ If $c=1$, this becomes $e^{-x}\sqrt{2\pi x}\to 0$, and if $c>1$ then $\frac{c^cx^{c-1}}{e^c}>1$ for $x\gg 0$ and hence $\left(\frac{c^cx^{c-1}}{e^c}\right)^{x}\sqrt{2\pi cx}$ grows faster tan $\sqrt{2\pi cx}\to \infty$.

Answer 2

No. The reason any number greater than 1 results in it approaching infinity is that as x approaches infinity, any constant in front of it greater than 1 will be multiplied by itself an infinite number of times. So any number greater than 1 will result in infinity.