Evaluate $\lim_{x \to \infty} \frac{(\frac x n)^x e^{-x}}{(x-2)!}$

Question

$$\lim_{x \to \infty} \frac{(\frac x n)^x e^{-x}}{(x-2)!}$$

where $x$ is $\mathbb N$-valued and $n$ is some nonzero real number.

Wolfram seems to give $0$ for different values of $n$ that I tried.

Is it really $0$ for any nonzero real $n$? I tried Stolz–Cesàro, but that doesn't seem to get me anywhere.

Answer 1

$$\require{cancel}\begin{align} \lim_{x \to \infty} \frac{\left(\frac x n\right)^x e^{-x}}{(x-2)!} &=\lim_{x\to\infty}\frac{x^x}{n^xe^x(x-2)!}\tag1\\ &=\lim_{x\to\infty}\frac{x^x(x-1)x}{n^x e^x \color{brown}{x!}}\tag2\\ &=\lim_{x\to\infty}\frac{x^x(x-1)x}{n^x e^x \color{brown}{\sqrt{2\pi x}\left(\tfrac xe\right)^x}}\tag3\\ &=\lim_{x\to\infty}\frac{\color{#C00}{x^x}(x-1)x\cdot {\color{royalblue}{e^x}}}{n^x \color{royalblue}{e^x} \sqrt{2\pi x}\color{#C00}{x^x}}\tag4\\ &=\lim_{x\to\infty}\frac{(x-1)x}{n^x\sqrt{2\pi x}}\tag5\\ &=\lim_{x\to\infty}\frac1{\sqrt{2\pi}}\frac{x^{3/2}-x^{1/2}}{n^x}\tag6\\ &=\begin{cases}\infty,&n\in(0,1]\\ 0,&n\in(1,\infty)\end{cases}.\tag7 \end{align}$$ Explanation:
$(2)$ $(x-2)!=\tfrac{x!}{(x-1)x}$.
$(3)$ Stirling's formula: $x! \sim \sqrt{2 \pi x}\left(\frac{x}{e}\right)^x$.
$(7)$ $n^x$ is strictly decreasing for $n\in(0,1)$ and strictly increasing for $n\in(1,\infty)$, in the former case the exponential grows much faster than $x\mapsto x^{3/2}-x^{1/2}$.