Answer:
We want to find:
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D)
Here we can use Stirling's approximation, which says that for large values of n, we get:

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]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7B%5Csqrt%7B2%2A%5Cpi%2An%7D%20%2A%28%5Cfrac%7Bn%7D%7Be%7D%20%29%5En%7D%20%7D%7Bn%7D%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7Bn%7D%7Be%2An%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D)
Now we can just simplify this, so we get:
![\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Be%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D%20%5C%5C)
And we can rewrite it as:

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus:

Answer:
Step-by-step explanation:
The zeros are the values of x for which y = 0
Quadratic formula
x = [5 ± √(5² – 4·1·6)] / [2·1]
= [5 ± √1] / 2
= [5 ± 1] /2
= 2, 3
sum of zeros = 5
Step by step explanation
4n-8=5
4n= 5+8
4n=13
4/4 n = 13/4
n= 13/4
Answer:15
Step-by-step explanation:5 and 3 both go into 15