Answer:
We want to find:
![\lim_(n \to \infty) \frac{\sqrt[n]{n!} }{n}](https://img.qammunity.org/2022/formulas/mathematics/college/n2nb5lpx29fqkviwhilxdl4b5jkaoqgn4e.png)
Here we can use Stirling's approximation, which says that for large values of n, we get:
![n! = √(2*\pi*n) *((n)/(e) )^n](https://img.qammunity.org/2022/formulas/mathematics/college/e0xpj0ok6lopywz5i3qtvx9dbdy4rolo0g.png)
Because here we are taking the limit when n tends to infinity, we can use this approximation.
Then we get.
![\lim_(n \to \infty) \frac{\sqrt[n]{n!} }{n} = \lim_(n \to \infty) \frac{\sqrt[n]{√(2*\pi*n) *((n)/(e) )^n} }{n} = \lim_(n \to \infty) (n)/(e*n) *\sqrt[2*n]{2*\pi*n}](https://img.qammunity.org/2022/formulas/mathematics/college/mahzipj1fd5bqurju573q6mc69m5xcw90b.png)
Now we can just simplify this, so we get:
![\lim_(n \to \infty) (1)/(e) *\sqrt[2*n]{2*\pi*n} \\](https://img.qammunity.org/2022/formulas/mathematics/college/3r6ch0nu8zz1aw5sp1g6chb237d2319sp2.png)
And we can rewrite it as:
![\lim_(n \to \infty) (1)/(e) *(2*\pi*n)^(1/2n)](https://img.qammunity.org/2022/formulas/mathematics/college/9we7zinzp52vizywgfxpayge1ng0doe6ph.png)
The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus:
![\lim_(n \to \infty) (1)/(e) *(2*\pi*n)^(1/2n) = (1)/(e)*1 = (1)/(e)](https://img.qammunity.org/2022/formulas/mathematics/college/50306rfx298x3ovo0iwsdlf0tluuz8hv8h.png)