Limit of x! over x^x as x goes to infinity

Limit of x!/x^x as x goes to infinity using the squeeze theorem.

[GermanAndres]

It's amazing how passionate you are about teaching, thank you!

Also "those who stop learning stop living" hit me hard.

[enirgetec]

Always a joy to watch these as a calc student who is beyond bored by my textbook's bland problems. Your passion is contagious, and you help me realize how beautiful math is. Thank you for these videos.

[gunz-oh]

I'm Japanese, and I'm not good at English so much, but your explanation is very easy to understand for me.
Thank you! and Excellent!

[adwz]

The Stirling approximation for the factorial would make this really quick!

x! ≡ Γ(x+1) ~ sqrt(2πx) (x/e)^x as x -> ∞

So x! / x^x = O(x^1/2 e^-x) as x -> ∞

-> 0 as x -> ∞, since e^-x is beyond all orders in x^k as x -> ∞

(and nice thing is this clearly holds for non-integer x via the gamma function, so no need to worry about the factorial of a non-integer aspect in a continuous limit)

[jamesharmon]

What I immediately looked at was how the numerator and denominator are defined. The numerator is 1*2*3*4...x.
.., the denominator is defined x*x*x*x...

Clearly, the denominator is getting bigger faster than the numerator, so the limit will be zero.

[Tmplar]

In Italy we call it “The Cops Theorem” because the two external functions are like cops carrying the middle function to their same limit (prison).

[cdkslakkend]

Honestly, great presentation. I understood from beginning to end. It's never always clear how creative logic can be applied when using inequalities. This example using the squeeze theorem to demonstrate how to rewrite the question in a form that looks much more digestible is priceless. Thank you.

[keithnisbet]

Wow. I haven't done this kind of math since 1972 when in Physics program at university. Long unused but not totally forgotten. You have a wonderful teaching style, far better than the "Professors" that I had at the time. I love the logic and reasoning that allows such seemingly difficult problems to be solved. Thanks very much.

[jayniesgottagun]

I never took Calc and I understood everything you said. You are marvelous.

[josearmandorz]

That is the most beautiful and satisfying limit demonstration I’ve ever seen

[tsulong]

I've been out of school for ~15 years and i don't use anything more advanced than basic algebra for my current job. Coming back to these concepts is so much fun and so interesting. And you're such a great teacher too!

[markrobinson]

From one math teacher to another, you are a great teacher.

[robvdm]

This is a nice demonstration of the kind of fundamentals that mathematicians use frequently that many students don’t really encounter. I do bounds and rate stuff quite frequently and there’s always a bunch of little tricks that I use to get things into a nice form that aren’t really “advanced” but also aren’t exactly easy. You need to have a good mathematical awareness for this kind of stuff.

[gallium-gonzollium]

honestly your presentation is so intuitive and awesome that i would want to have you as my calculus teacher. no joke youre actually on par with 3blue1brown, if not beyond, when it comes to visual learning like this. i commend the phenomal work here.

[iceify]

Bro, I just found this channel, and this is really great stuff. This wasnt new, yet very plainly explained. Great to see that the math content creators are not 100% whitebread

[alexandre]

Haha great video! Studied calculus 20 years ago, I can still follow you... I'm glad I put the effort into learning it at the time! Thanks for the video, you made it look easy!

[jimmybee]

For every x >= 2, x! is smaller than x^x BECAUSE

x^x = x•x•x•x…x•x (x times)
x! = x•(x-1)•(x-2)…(2)•(1)

The terms of x! are getting farther away from x, so x^x would in a way reach infinity faster, so the expression is like (small infinity)/(big infinity). This is more easily seen as 1/(infinity) or just 0.

*Also multiplying out x! gives some polynomial with leading coefficient one: x^x - x^x + (xC2)x^(x-2) + …

This means the degree of the numerator is smaller than the degree of the denominator, so the limit is zero.

(x choose 2)x^(x-2) + …
—————————————
x^x

I think..?

[mikezilberbrand]

then the ratio is 1/x* Each one is <or=1. Thus the lim < lim(1/x)=0

[foobar]

Does anyone else feel that there is sleight of hand in using <= rather than < to make the final result look nicer? After all, we are considering x approaching infinity and the values of x where there is equality is small and known - i.e. 1 and possibly 0. I guess saying a < b < c where a and c approach 0 means b approaches 0 is valid but does not look quite so neat.

[alikaperdue]

My initial guess by looking at it is that it will approach 0. Because breaking it apart it will be a lot of factor terms that start with finites (1, 2, 3...) on the top and infinite on the bottom. Leading up to factors that approach 1.