Why is the limit of (factorial of n) divided by (x power of E) (x tends to be positive infinity) infinitesimal (factorial of n) is the numerator, (x power of E) is the denominator X - > positive infinity N belongs to integer Z

Why is the limit of (factorial of n) divided by (x power of E) (x tends to be positive infinity) infinitesimal (factorial of n) is the numerator, (x power of E) is the denominator X - > positive infinity N belongs to integer Z

When x tends to positive infinity, the x power of e tends to positive infinity, and the factorial of n is a constant
So the limit is infinitesimal

The product of the n-th power of 2 and the factorial of N divided by the limit of the n-th power of n when n tends to infinity

Using Stirling's formula! (2 * pi * n) ^ 0.5 * (n / E) ^ n (PI is the PI and E is the base of the natural logarithm (Euler constant)), so Lim 2 ^ n * n/ N ^ n = Lim 2 ^ n * (2 * pi * n) ^ 0.5 * (n / E) ^ n / N ^ n = (2 * pi * n) ^ 0.5 * 2 ^ n / e ^ n because E > 2, the limit is

Using the definition of limit, it is proved that the order multiplication and division of n is to the nth power of n

The establishment of the first floor also requires that the limit of (n / N) * [(n-1) / N] * [(n-2) / N] *... Is limited
It should be like this 1 / (n ^ n) / N= 1/(n/1*n/2*n/3*.*n/n)
It can be obtained that all factors of N / 1 * n / 2 * n / 3 *. * n / N are greater than 1 and greater than N, and the limit is infinite, so the limit of 1 / (n / 1 * n / 2 * n / 3 *. * n / n) is 0

If the power a of 2 is equal to the power B of 5 is equal to 10, how much is one part a plus one part B?

There are two ways to solve this problem:
First kind
2^a=5^b=10
alg2=blg5=1
a=1/lg2
b=1/lg5
1/a+1/b
=lg2+lg5
=lg10
=1
Second:
The power a of 2 is equal to the power B of 5 = 10
Then 5 ^ (A-1) = 2
2^(b-1)=5
5 ^ (A-1) = 2, take the power of (B-1) on both sides
Then: 5 ^ (A-1) (B-1) = 2 ^ (B-1)
Then: 5 ^ (A-1) (B-1) = 5
So: (A-1) (B-1) = 1
Namely: (a + b) / AB = 1
So: 1 / A + 1 / b = 1

The power a of 3 = the power B of 5, so as to know that the power a of 3 = the power B of 5 = a, and one of a + one of B = 2, then a is equal to?

Because 3 ^ a = 5 ^ B = a 1 / A + 1 / b = 2
So first take all the logarithms on that formula
aln3=bln5=lnc
1/a=ln3/lnc
1/b=ln5/lnc
So there are
(ln3+ln5)/lnc=2
lnc=1/2ln15
c=√15

If a is equal to the 99th power of 9 and B is equal to the 99th power of 9 and the 9th power of 11 / 90th power of 9, is a greater than B

a=99⁹/9⁹⁹
=9⁹ × 11⁹/9⁹⁹
=11⁹/9⁹⁹
b=11⁹/9⁹⁰
So a = B