# Misc Solution: An interesting Limit

Solve the following limit with the nth root of the factorial divided by $$n$$:

$L = \lim\limits_{n\to\infty} \frac{\sqrt[n]{n!}}{n} = \lim\limits_{n\to\infty}\left(\frac{n!}{n^n}\right)^\frac{1}{n}$

## Solution

The problem is that $$\lim\limits_{n\to\infty}\frac{n!}{n^n}=0$$ as well as $$\lim\limits_{n\to\infty}\frac{1}{n}=0$$, which leads to $$0^0$$ and thus can have an arbitrary solution. However, when we take the log on both sides, we get

$\begin{array}{rl} \log(L) &= \log\left(\lim\limits_{n\to\infty}\left(\frac{n!}{n^n}\right)^\frac{1}{n}\right)\\ &= \lim\limits_{n\to\infty}\left(\log\left(\frac{n!}{n^n}\right)^\frac{1}{n}\right)\\ &= \lim\limits_{n\to\infty}\frac{1}{n}\log\left(\frac{n!}{n^n}\right)\\ &= \lim\limits_{n\to\infty}\frac{1}{n}\left(\log\left(\frac{1}{n}\right) + \log\left(\frac{2}{n}\right) + \dots+ \log\left(\frac{n-1}{n}\right) + \log\left(\frac{n}{n}\right)\right) \end{array}$

Now remembering the definition of the Riemann sum, with which we can approximate the area under a function $$f(x)$$ in the intervall $$a$$ to $$b$$

$\lim\limits_{n\to\infty}\sum\limits_{i=1}^nf(x_i)\cdot\Delta x_i = \int\limits_a^bf(x)dx$

we can solve for the log of $$L$$

$\begin{array}{rl} \log(L) &= \lim\limits_{n\to\infty}\sum\limits_{i=1}^n \log\left(\frac{i}{n}\right)\cdot \frac{1}{n}\\ &= \int\limits_0^1\log(x)dx\\ &= x\cdot\log(x) - x\Bigm|_0^1\\ &= (1\cdot\log(1) - 1) - (\underset{0\cdot\log(0)}{\to 0} - 0)\\ &= -1 \end{array}$

And therefore solving for $$L$$:

$\begin{array}{rl} L &= \lim\limits_{n\to\infty}\left(\frac{n!}{n^n}\right)^\frac{1}{n}\\ &= e^{\log(L)} \\ &= e^{-1} \\ &= \frac{1}{e} \end{array}$

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