Limit $(n^x)/x!$ as $x$ tends to $\infty$?

Question

I can't seem to derive $$\lim_{x \rightarrow \infty} \frac{n^{x}}{x!}$$ Any help appreciated

Answer 1

Since the Taylor Series for the exponential function, $$e^n=\sum_{x=0}^{\infty}\frac{n^x}{x!}$$ converges for any $n\in\Bbb R$, what can you say about $\lim_{x\to\infty}\frac{n^x}{x!}$?

Answer 2

HINT:

Show that $x!\ge (x/e)^{x}$. Hence, for $x>2en$

$$\frac{n^x}{x!}\le\frac{n^x}{(x/e)^x}=\left(\frac{en}{x}\right)^x\le \frac{1}{2^x}$$

SPOILER ALERT: Scroll over the highlighted area to reveal the solution

We will assume that $x\in \mathbb{N}$. This is not a restriction inasmuch as for $x\in \mathbb{R}$, $x!=\Gamma(x-1)$, where $\Gamma(x)$ is the Gamma function. To show that $x!\ge (x/e)$, we write $$\begin{align}\log(x!)&= \sum_{k=1}^{x}\log(k)\\\\&=\sum_{k=1}^{x}\log(k/x)+x\log(x)\\\\&=x\log(x)+x\underbrace{\left(\frac{1}{x}\sum_{k=1}^{x}\log(k/x)\right)}_{\text{Riemann Sum for }\,\,\int_0^1 \log(x)\,dx=-1}\\\\&\ge x\log(x/e)-x\\\\&=\log\left((x/e)^x\right) \end{align}$$whence we have the inequality $x!\ge (x/e)^x$ as was to be shown.