Why does $\lim_{x\to\infty} \frac{x!}{x^x}=0$?

Question

I'm in calc 2 and we're doing series right now. I was using the alternating series test and ran into this: $$\lim_{x\to\infty} \frac{x!}{x^x}$$ I don't understand how the limit is $0$. The numerator would end up larger than the denominator, but why does that make the limit $0$? Is it that the numerator approaching infinity is so large that the denominator might as well be $0$? But you can't divide by $0$ so I don't get it. Maybe I'm missing a super basic limit law right now...

Answer 1

From $\displaystyle 0\leq \frac{x!}{x^x}=\frac{x}{x}\cdot \frac{(x-1)}{x}\cdot \frac{(x-2)}{x}\cdots\frac{1}{x}<1\cdot 1\cdot \cdots \frac{1}{x}. $

Now applying limit $x\rightarrow \infty$ and using Squeeze theorem , We get $$\lim_{x\rightarrow \infty}\frac{x!}{x^x}=0$$

Answer 2

Solving the limit: $$ L =\lim_{x \to \infty} \frac{x!}{x^{x}}$$ apply logarithm both sides $$ \ln L=\lim_{x \to \infty} \ln {x!} - x\ln {x}$$ By Stirling's approx. when $x \to \infty$ $$\ln {x!} = x\ln {x} -x$$ So therefore, $$ \ln L=\lim_{x \to \infty} \ x\ln {x} -x - x\ln {x}$$ $$ \ln L=\lim_{x \to \infty} -x$$ $$ L = \lim_{x \to \infty} e^{-x}$$ $$ L = 0$$

Edit1: Solving it without Stirling's approx.(I changed the variables so it is easy for me to write the answer) $$ \ln L=\lim_{n \to \infty} \ln {n!} - n\ln {n}$$ After this we can write it as a limit of a sum $$ \ln L=\lim_{n \to \infty} \left(\sum_{k=1}^{n}\ln {k!}\right) - n\ln {n}$$ $$\sum_{k=1}^{n} \ln {k!} \approx \int_{1}^{n} \ln {x}dx = n\ln {n} - n + 1$$ The above statement is only true when $n \to \infty$ Now the limit becomes $$ \ln L=\lim_{n \to \infty} n\ln {n} - n + 1 - n\ln {n}$$ $$ \ln L=\lim_{n \to \infty} -n+1$$ $$ L = \lim_{n \to \infty} e^{-n+1}$$ $$ L = 0$$

Answer 3

Think of $x$ balls.

1. The # of arrangements of $x$ balls $x$ at a time when repetition is not allowed = $^xP_x = x!$
2. The # of arrangements of $x$ balls $x$ at a time when repetition is allowed = $x^x$

Which is bigger?

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$x! = x(x - 1)(x - 2) ... 3 \cdot 2 \cdot 1$

$x - k < x$ where $k$ is any number and so if $x^p = x - 1, p < 1 \implies p = 1 - d$

So, $x! = x^1 \times x^{1 - d_1} \times x^{1 - d_2} \times .... \times x^{1 - d_{x - 1}}$

$x! = x^{x - (d_1 + d_2 + ... + d_{x - 1})}$

Let $d_1 + d_2 + ... + d_{x - 1} = s$

So, $\frac{x!}{x^x} = \frac{x^{x - s}}{x^x} = \frac{1}{x^s}$

$\displaystyle \lim_{x \to \infty} \frac{x!}{x^x} = \lim_{x \to \infty} \frac{1}{x^s} = 0$

Answer 4

A proof using only rational numbers (I assume that $x\in\Bbb Z_+$):

For $x\ge 6$ (for example), let $k$ be the floor of $x/2$. Then $$\frac{x!}{x^x}\le \underbrace{\frac{x/2}x\cdot\frac{x/2}x\cdot\ldots\cdot\frac{x/2}x}_{k\text{ times}}\cdot \underbrace{\frac xx\cdot\frac xx\cdot\ldots\cdot\frac xx}_{x-k \text{ times}} = \left(\frac12\right)^k\le\left(\frac{25}{49}\right)^k\le\left(\frac {5}{7}\right)^x$$

Answer 5

Let $a_n=\frac{n!}{n^n}$. Then

$$\ln a_n=\ln n!-n\ln n=\sum_{j=1}^n(\ln j-\ln n)=\sum_{j=1}^n\ln\frac{j}{n}\leq \ln\frac{1}{n}\to-\infty$$

as $n\to\infty$, where the inequality comes from the fact that $\ln\frac{j}{n}\leq0$ for all $j\in\{2,\dots,n\}$. As $e^x\to0$ as $x\to-\infty$, it follows that $a_n\to0$ as $n\to\infty$. Note that we we don't take $n$ to be a natural number, but instead a positive real, the result follows from this, as the function $x\mapsto\frac{\Gamma(x+1)}{x^x}$ is decreasing for sufficiently large $x$.