What is infinity factorial over infinity power infinity

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It has no meaning for infinity is not a number but just a conception! Factorial works only for integers bigger than zero, despite 0! =1 with some forcing. Furthermore, infinity powered to infinity is a indefinition and must be treated within calculus not as if it were a number but being a limit. I admire your effort to bring a curious question but those concepts only apply to real numbers!( and factorial just to integers).

GIFPES

Just as a vertical line has no slope while a horizontal line has a zero slope, similarly, infinity has no numerical value whereas zero has a zero value.

stevenmayhew

I'd conclude, without formal proof, that [infinity^infinity] Is so large that the fraction becomes 0.
As a guess at solution -->

vs
[infin!]/[infin^infin]
yields results

tomtke

How did you deduce Equation 1 is greater than 1?

Should It not be lesser than 1?

o-

If you accept transfinite cardinals as a valid contruct, then the numerator would be aleph zero while the denominator would be 2 to the power of aleph zero. This expression would thus appear to equate to zero since the numerator is infinitesimally small compared to the denominator. This assumes that the infinity symbol in this context is equivalent to the cardinal aleph zero. In mathematics, while the extended reals are shown as including the symbols plus and minus infinity, it is a mistake to think that the symbol is a number and treat it as such. Its inclusion as a symbol allows Dedekind-MacNeille completeness which allows our expression of limits et alia. Infinity is not a real number.

paulstclairterry

Infinity to the infinity power accelerates at a faster rate than infinity factorial.

josephrusso

Instead of X! let's use Stirling's formula
x! ≈ sqrt(2πx)*(x/e)ˣ
then
x!/xˣ ≈ sqrt(2πx)/eˣ
(x->∞) so lim[sqrt(2πx)/eˣ] = lim[sqrt(2π)/(2sqrt(x)*eˣ)] = 0

This is not proof, it is a quick estimate that you can do in your head.

grzegorzkondracki

As written it's undefined. As a limit problem where all the infinity symbols shown are xs approaching infinity it's zero. I can start writing 1*2*3 but start replacing the rest with xs and it becomes a lower power of xs over a higher power of xs which is zero.

KingGisInDaHouse

according to the mathematical definition of infinity, any action with it in the numerator will be equal to infinity. This is not a task, but a compulsion to tautology.

fedorovd

3:25 Why x/x is "less or equal" 1. How it can be less than 1? It is strictly =1

ORTHODOXLOGOS

Think you missed a factor in x exp x: @1, x! = x/x exp 1, for 2, x! = 2. whereas x exp x = x exp 2, and the series becomes 2x/x exp 2. The thrid term would be x3/x exp 3, 3x/x exp 3 in a continuing declining series approaching zero. I do agree that using "infinity" is fundamentally undefined, but ... thought provoking!

stevereade

Obviously, Infinity to the infinite power is still infinity. And the same with infinity factorial.

MikeDanielsen

The native question is ambiguous, thus the answer is : ' ? ', due to insufficient input . CONCLUSION : the native answer ' ? ' is perfect .

martinlutherwrong

so 0!/0^0=0 it means that 1/0^0=0 which also means that 0^0=1/0 which also means that 0*0^0=1 and that also means that 0*0^0=0!

bozydarziemniak

lim n→∞ (n! / n^n) = lim n→∞ Π(1/n) = 0

yomiya-ch

the answer is correct
lim(x−>∞) (x!/x^x) =0
but the question is wrong as infinity is not a number.

davidseed

Стремится к 0 и является бесконечно малым. Это очевидно и без вычислений посколтку х^х растет быстрее чем х!

AlCapone-gd

Are we dealing with countably infinite? Or a higher order of infinity? I took set theory a long time ago, so I'm a little rusty, but I remember one-to-one correspondence and the diagonalization proof.

dsperalta

What does it "(i) and (ii)" mean ?

ryszardmkrak

Infinity is not a number. It is a concept.
You cannot substitute values for the infinity character.

The limit x -> infinity (x!/x^x) is correct.
But substituting x with the infinity character isn't.

wimahlers

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