Calculus 2: List the functions below in order of size from smallest to largest

If n is sufficiently large, the following functions of n canbe arranged in an increasing
order, so that each function is very much larger than the onepreceding it. List the functions
below in order of size from smallest to largest.
n ; nn ; ln n ;4n ; 2n ; n ln n ; 2(n^2) ; sqrt(n6 + 1) ; (n3 + 1)2/3 ; n!

PlEASE HELP, TEN POINTS FOR BEST ANSWER, YOUR HELP IS GREATLY APRRECIATED! :)

For n very large,

(n^3 + 1)^(2/3) behaves like (n^3)^(2/3) = n².

Similarly,

√(n^6 + 1) behaves like √(n^6) = n^3.

It is pretty easy to classify these two relative to each other. It also helps to know that exponentials dominate powers, and powers dominate logs. You can also use the fact that f(n) dominates g(n) if

lim g(n)/f(n) = 0.
n->∞

I'm still not sure if I understand all of the functions in your sequence. Are they

n, n^n, ln(n), 4ⁿ, 2ⁿ, n ln(n), 2n², √(n^6 + 1), (n^3 + 1)^(3/2), and n!

or just the five functions in the additional details?

Oh well, I'll order them all, and you can throw out any you don't need.

ln(n) is the smallest, and n is next, followed by n ln(n).

Next come regular powers (not n^n that is different).

(n^3 + 1)^(3/2), 2n², √(n^6 + 1) ---- these are like n², 2n², n^3

Next come the regular exponentials (not n^n it is still different)

2ⁿ, 4ⁿ

Then finally n! and lastly n^n.

To see why n^n is bigger than n!, remember that n! is 1•2•3•••n. This is the product of n numbers where all but the last is smaller than n. By contrast n^n = n•n•n•••n (n times) is the product of n numbers that are all equal to n.

In conclusion

ln(n), n, nln(n), (n^3 + 1)^(2/3), 2n², √(n^6 + 1), 2ⁿ, 4ⁿ, n!, n^n