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

Prime Newtons

5 місяців тому

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

КОМЕНТАРІ: 611

@GermanAndres 5 місяців тому

It's amazing how passionate you are about teaching, thank you! Also "those who stop learning stop living" hit me hard.

@harriehausenman8623 Місяць тому

This channel is a true gem.

@Tmplar 5 місяців тому

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).

@whannabi Місяць тому

Same in France except we don't say cops but "gendarme" which is different but for the sake simplicity let's just say they're a kind of cop.

@jamesharmon4994 5 місяців тому

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.

@rodbenson5879 5 місяців тому

Yes this one is very simple.

@zz-nc5kx 5 місяців тому

Yup! I quickly came to the same conclusion.

@adw1z 5 місяців тому

It’s still not obvious that the limit tends to 0 and not some constant in (0,1) - that requires proving

@zz-nc5kx 5 місяців тому

@@adw1z pretty obvious to me, especially when observing x! And x^x using my old friend Desmos. For x>0, x^x clearly blows away x!.

@christie6279 5 місяців тому

why does the denom increasing faster than the numerator mean the limit is 0?

@gunz-oh3917 4 місяці тому

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!

@PrimeNewtons 4 місяці тому

Glad to hear that!

@2sljmath 3 місяці тому

👌🏻👌🏻

@enirgetec3523 5 місяців тому

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.

@ayushrudra8600 5 місяців тому

You should use aops books if you want more interesting problems / more of a challenge

@Kaze1111 2 місяці тому

What’s calc? UC Berkeley? They should have something way much harder than this

@cdkslakkend5742 5 місяців тому

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.

@robvdm 5 місяців тому

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.

@keithnisbet Місяць тому

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.

@josearmandorz75 Місяць тому

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

@markrobinson9956 5 місяців тому

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

@harriehausenman8623 Місяць тому

Such a beautiful chalkboard-writing 🤗 It really helps digesting the content since it is all so clearly readable and nicely ordered. Compared to others (*cough*Borcherds*cough*) this should have waaay more views and subs! Thanks for the great content!

@alexandre9051 5 місяців тому

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!

@gallium-gonzollium 5 місяців тому

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.

@juv7026 2 місяці тому

..what?

@adw1z 5 місяців тому

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)

@scottparkins1634 5 місяців тому

Yep that’s true, you can also use the AM-GM to get an upper bound of [(x+1)/2]^x for x! but this is overkill compared to the simple method presented here 😂

@maelhostettler1004 5 місяців тому

however the proof of stirling involve Wallis Integral and general properties of equivalents... not that ez

@adw1z 5 місяців тому

@@maelhostettler1004 I’ve done exactly that so I am satisfied, not sure about everyone else though 😭 Also proved the Stirling series to the next order term using Laplace’s Method and And Watson’s Lemma, in fact: x! ≡ Γ(x+1) ~ sqrt(2πx) (x/e)^x * [ 1 + 1/(12x) + 1/(288x^2) + … ] as x -> ∞ The first term of the series is the commonly known “Stirling Approximation”, which in itself is extremely accurate for large x so the other terms aren’t really needed, but still it’s very interesting! The point is to use what we already know - there’s no point in not using the Stirling Formula given it’s been proved and is now a common result. It’s like saying you can’t use the squeeze theorem, because u don’t know the proof of it. The result I showed is better anyways as it shows the asymptotic behaviour of the limit function at large x precisely, and not just the limiting result at infinity. And arguably, the Stirling series is not hard to show - just need a few integration by parts and clever substitutions. The foundations are set in stone

@lgooch 5 місяців тому

@@scottparkins1634I don’t think this is overkill, this a nice solution.

@filippomariachiappini1257 5 місяців тому

@@maelhostettler1004very important observation! We often think some proofs are faster or simpler when they actually require some more advanced techniques, that require some longer or more difficult proofs. The beauty of mathematics is to prove apparently difficult statements with elementary techniques, trying to not overcomplicate things.

@inyobill 5 місяців тому

Good stuff, mate. Super clear discussion. Making complex concepts easy is a gift. I see why you're on the way to your million sub=scribers.

@tsulong 5 місяців тому

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!

@knupug 5 місяців тому

I've been out of high school for 51 years (calc 1 and 2) and college for 44 (diff eq) and couldn't agree more!!

@indescribablecardinal6571 5 місяців тому

And that's sad. Research, programming with luck and teaching are the only accesible jobs where you can apply advanced maths.

@jayniesgottagun 5 місяців тому

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

@KSM94K 5 місяців тому

That's big brain

@user-fm1sr2fu3z Місяць тому

One of the best videos I've ever seen. Fascinating problem solved in a fundamental yet brilliant way!!! Keep up the great work

@alikaperdue 5 місяців тому

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.

@GigaChad46 2 місяці тому

I thought it too. Because x! is slower growing than x^x, thus even for small intigers making patern: Let x=3 3!/3³ = 6/27 = 2/9 Let x=4 4!/4⁴ = 24/256 = 3/32 Since 2/9 > 3/32, we can say this fuction tends to go to zero.

@ice9ify 5 місяців тому

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

@Xenomnipotent 4 місяці тому

Just want to say I absolutely love your videos! Your energy and enthusiasm are so captivating and really makes me appreciate mathematics much more.

@Frostnburn 5 місяців тому

For positive integer X, we know that X! = X * (X-1) * (X-2) * .... * 1, totalling X terms, we know X^X is X * X * X... * X, X terms, so X!/X^X is X/X * (X-1)/X * (X-2)/X * ... * 1/X We know that the first term is 1, the second term onwards all the way to 1/X is less than 1, so the original function is always less than 1 for any positive integer X. As X approaches infinity, the last term (1/X) approaches zero, so the original function must also approach zero.

@BossDropbear 5 місяців тому

Exactly. Not sure how this is a 10 min video.

@emurphy42 5 місяців тому

@@BossDropbearWhat would be more interesting is proving the same for the gamma function (generalizing factorial from integers to reals). Intuitively the ratio probably still continuously decreases, but I haven't tried to prove that it does.

@slr150 Місяць тому

Yeah I did the same thing, but It's interesting to see other approaches.

@gultuteferra8342 4 місяці тому

I’ve never seen such a best teacher. Thank you so much.

@foobar476 5 місяців тому

Does anyone else feel that there is sleight of hand in using

@user-qd6hx7yh5o 4 місяці тому

It's an amazing approach! Thank you from Russia!

@jimmybee7 5 місяців тому

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..?

@DeadCatX2 5 місяців тому

This was my immediate intuition as well, the denominator "grows" faster so the result should approach zero as we approach infinity

@rodbenson5879 5 місяців тому

Yep.

@wiilli4471 5 місяців тому

This is not a mathematical proof lmao.

@sigsqrl 5 місяців тому

⁠@@wiilli4471did they say it is?

@Bodyknock 5 місяців тому

It’s not quite that simple. For instance, consider the functions f(n) = (1/2 + 1/2ⁿ) n and g(n) = n, and look at f/g. The numerator decreases over time approaching n/2 while the denominator is always n. Notice that f is always moving “farther away” from the denominator, but their ratio is approaching 1/2 and not zero.

@nevvermind75 4 місяці тому

I was very much enjoying the video, but the outro got you a new subscriber. You have a very theatrical & charismatic way of talking. I love it!

@adriagonzalezroige1337 5 місяців тому

Dude, I've never seen you before, one minute into the video and I can see how passionate you are about math, I love it dude! Have a great day

@KennyMccormicklul 5 місяців тому

man i tell you.. i dont get many new things that i listen to at the first time.. but in this case, i understood it at the first time. thank you man you are great :D!

@PrimeNewtons 5 місяців тому

Glad to hear that!

@johnfranchina84 Місяць тому

Electrical Engineer here - my degree was like a deep dive into maths which I loved. Love your passion for maths snd teaching.

@Sentient_Blob 5 місяців тому

Really cool video! We touched on this concept in calc but never really used it, it’s nice to see it applied

@utuberaj60 5 місяців тому

Absolutely great Mr. Newton. However for positive integers we can easily see by inspection that the numerator will have the highest power of X as (X-1), whereas the denominator is X^X- that will be simply an expression with 1/X ---> infinity gives the answer = 0

@bradryan8071 3 місяці тому

An alternate approach is to use the Limit of a product is the product of the limits. Thus the limit becomes ( lim x/x ) ( lim x-1/x )( lim x-2/x) ... (lim x/2)(lim x/1 ) . The first limit as x goes to infinity is one, the rest in the line of products is zero. Therefore the limit is zero.

@eng954 Місяць тому

your explanation and english are both very clear and understandable. As an old mentor of engineering math. i appreciated you so much.Thank u so much.

@worldnotworld 3 місяці тому

Very clear. And I love your cap. It suits you!

@DiegoAndrade 5 місяців тому

Bravo beautiful introduction keep it coming brother !!!

@sugaruisland6387 4 місяці тому

I don't think "the squeeze theorem" is easy, but your neat and gentle explanation makes me understand this theorem. Even I'm not native speaker of English. Thank you very much.

@KarlChamoun 5 місяців тому

just a clarification, the gamma function can be defined for negative non integers and can also be lower than 1 for positive numbers

@PrimeNewtons 5 місяців тому

True. I realized what I said but it was too late 😢

@Steve_Stowers 5 місяців тому

If we're assuming that x is a whole number, so that x! is defined, I would have written it with n, rather than x. Sometimes this doesn't make a difference, but sometimes it does. For example, the sequence sin(πn) is just 0, 0, 0, ... so it has limit 0. But the limit of the function sin(πx) (where x is a real number) as x approaches infinity does not exist.

@michaelandcarmenmaguire1108 5 місяців тому

Thank you for your passion. Very well presented proof.

@mikezilberbrand1663 2 місяці тому

X!=1*2*3*4*5.....*X, then the ratio is 1/x* 2/x*.....x/x. Each one is

@leif1075 Місяць тому

Can you clarify your notation please?

@holmbrg-_-2221 5 місяців тому

I havent heard about the squeeze theorem before, im not on that level yet i guess, but thats actually super useful. Thank you for this video.

@Dongobog-ps9tz 4 місяці тому

You have a delightful voice! I'd listen to you read an audiobook

@ap1962 Місяць тому

I really like your use and explanation of the squeeze theorom

@AlbertTheGamer-gk7sn 5 місяців тому

Well, the limit as x goes to 0 is 1, due to the Taylor polynomial of e^x, where e^0 = 1 and has a 0^0/0! term that equals to 1, meaning that its inverse, 0!/0^0 is also equal to 1.

@user-ed4kj8yx9y 2 місяці тому

You consider the series sum_n^{infty} n!/n^n, use the quotient test, conclude that the series converges and deduce the limit of the sequence of summands to be zero. Note that when you write x!, you would typically mean the (continuous) Gamma-function, which coincides with a factorial only for integers x.

@carlosalbertocuadros5469 5 місяців тому

Good Job Professor

@curtiswfranks 5 місяців тому

Dude, where are you from? Your voice is so smooth and soothing. Can you just narrate my whole life? I can feel my blood pressure dropping as I listen to you.

@Ukraine-is-Corrupt 5 місяців тому

According to MatCad, that question isn’t expressed in the correct way. You should define whether the limit is being approached from below, from above, or from both sides of infinity. Remember, it has been proven that different sizes of infinity exist

@oddwad6290 3 місяці тому

Very clever and well done . Enjoyed watching it simplified .

@shgysk8zer0 5 місяців тому

In this case it's obvious that x can't be negative, but in the general case how is the < ruled out to give just the = case?

@kevinferrin5695 5 місяців тому

So smooth. Thank you.

@AbouTaim-Lille 5 місяців тому

You don't need to use any theorem. Just rewrite x^x as x.x.x...x and X! as X(x-1)(x-2) .... 2.1. and it is not hard to see that the fraction is smaller than 1/X which tends to zero at the infinity. But we usually compare n^n together with n! Multiplied by a^n a>1. Which really makes sens. And to do so we know that if n is sufficiently large then after some fixed integer N, we look for the limit of α(n)/α(n+1). And it should be equal to one of three values ∞, 0 or some integer.

@DavidCaveperson 5 місяців тому

X!/(X^X) = (X-1)!/(X^(X-1)) = Product (a) where a

@beentheredonethatunfortunately Місяць тому

I got here by videos I think are aimed at middle schoolers. Even if the title screen made me feel I must be missing something in my basic understanding. I was always annoyed by the presentation and ended up just looking at the comments to check I was correct - and yes always was. However this is much more interesting. Made me actually think and I even learned (or with my memory might even be reminded of) something. I had the right answer but I certainly didn't write in down and do it rigourlessly I and if I had written it down I wouldn't have got marks for the working out section (basically I just went straight from x!/x^x to 1/x and didn't do any "bracketing" for the squeeze). Good video. I look forward to finding more from you in my feed.

@BlueSiege01 5 місяців тому

Thank you, Sir!

@gabrielmarino8510 5 місяців тому

Unmatched teaching, even knowning the theorem I would never had though in using it.

@TechnoCoderz369 5 місяців тому

Literally I used to think that squeeze theorem is useless! Thanks for this video!

@TheDigiWorld 5 місяців тому

A simple way is to use the definition of factorial and exponents x! = x * (x - 1) * (x - 2) ... 2 * 1 x^x = x * x * x * x ... (x times) Note when we compute x!/x^x, the first x matches, but all the remaining terms in the numerator will be less than the remaining terms in the denominator as we let x go out towards ♾️. Which means that the denominator grows much faster than the numerator thus the limit approaches 0

@jamesharmon4994 5 місяців тому

I loved your explanation!

@pk2712 5 місяців тому

Great explanation and use of the squeeze theorem.

@sherylbegby 5 місяців тому

Beautiful proof and great explanation. Instant subscribe.

@naghipakdaman 5 місяців тому

Your way of teaching is just amazinggg

@ganesanthenappan5366 5 місяців тому

Beautifully Explained. Thanks

@OpPhilo03 5 місяців тому

Amezing teaching style sir. I impressed you.😊😊

@XxKnuckleSOverlorDxX Місяць тому

| x!/x^x |< E, for any E > 0 by the archimedian principle you can always find x such that the inequality holds for any arbitrary E.

@jakubl8271 2 місяці тому

Artificial time inflating. For example (x^(x-1))/(x^x) can be simplified much faster: (x^(x-1))/(x^x) = (x^(x-1))/(x*x^(x-1)), then x^(x-1) cancel out ans 1/x is left.

@Mal1234567 2 місяці тому

Claude 2 used Stirling's approximation for the factorial function: x! ≈ sqrt(2πx) * (x/e)^x I've never heard of it but it apparently worked.

@henryparker4668 5 місяців тому

Very cool explanation! Thank you. In looking at this problem, in my mind, I figured the denominator (x^x) would approach infinity faster than the numerator (x!). With that thought, I “guessed” the limit would be zero. Is there any logic to that thought process?

@PrimeNewtons 5 місяців тому

Yes. Almost every limit that converges to 0 as x goes to infinity is predictable. The problem is the algebra.

@mokoufujiwara4209 5 місяців тому

If you talk about integer x You can do that by splitting into 1/x * 2/x * ... * x/x there are x terms, all of them ∞)(1/x * 2/x * ... * x/x) = 0*0*...*1 = 0 Some people have shown other methods too, all of them worth reading.

@m.h.6470 5 місяців тому

Solution without any calculation: x^x grows much faster than x!, because if you expand those Terms, for every term of x^x, you have a smaller term (except for the first, which is equal) in x!. As such, the Limit tends to 0, as the denominator becomes much larger than the numerator.

@daales. 5 місяців тому

I just thought of x factorial as x(x-1)(x-2) and so on while x^x is x*x*x, so x! is being multiplied by less each time meaning that it is increasing at a lower rate, or in other words the function is bottom heavy. so y=0 because it is bottom heavy

@hamzaemad8338 5 місяців тому

Thank you you opened a way in my mind in maths section the way of your solving is very logic and good

@BigOttomatic Місяць тому

This was a nice way to show not only how to use the squeeze theorem, but also why it works

@willa4you 3 місяці тому

Fun fact: in Italy we call the squeeze theorem, the "teorema dei 2 carabinieri". Now, the carabinieri are technically policemen so the allegory is that two policemen heading somewhere are dragging with them the central function which is some kind of prisoner! 🙂

@PrimeNewtons 3 місяці тому

That's a beautiful allegory.

@MouhibBayounes 2 місяці тому

In french too!! We call it "theoreme des gens d'armes" 😂

@Rhiannon_Autumn 2 місяці тому

my explanation of why the limit equals zero, not even near as rigorous was that x^x grows in magnitude quicker than x! which was easily proved by taking a few exmples, so as x tends to infinity the numerator value is much smaller than the denominator, and becomes closer and closer to zero compared to the denominator as x increases even more. so the limit is zero over infinity which is zero.

@5anatakos 4 місяці тому

For non-integer values of x, one cannot say that x! = x*(x-1)*(x-2)*...*1 so, yes, one will need to use the Gamma function.

@donsena2013 Місяць тому

We could simply say that by mathematical induction 1 x 2 x 3 x 4 x 5 x . . . x n-1 x n = n ! is necessarily less than n x n x n x n x n x . . . x n-1 x n = X^X since, while there are n multipliers in each product, each multiplier in the denominator for the expression lim[X→ ∞] (X! /X^X) is less than or just equal to n. The significance of this disparity between the corresponding multipliers in X! and in X^X grows as the number of multipliers in X! and X^X itself grows, so that, for increasingly larger X, X! /X^X necessarily tends to zero.

@holdenmccrotch6485 5 місяців тому

Very clever. I thought I'd be too ignorant to understand this but I'm pretty sure I got it! Thank you for that

@billthomas7644 5 місяців тому

Thanks for the clear explanation.

@AcademiaCS1 4 місяці тому

That's it. The way you teach is really great. You allow (as it should be) students to think before you pronounce the answer, and that's the way we work at my Academy. I look up to you. :-)

@unclesmrgol 5 місяців тому

Best use of the squeeze theorem I've seen in a very long time.

@rezamalihi7120 5 місяців тому

Thank you, Sir! Great explanation!

@kokopelli314 Місяць тому

The fact that you could figure that out without even writing it down makes me happy

@justarandomdood 5 місяців тому

Isnt this just growth rate of functions? Ln(x) < x^n < n^x < x! < x^x Or something like that?

@davidconlee2196 5 місяців тому

Very cool! Thank you for sharing this

@banan0505 5 місяців тому

im glued to your channel now

@tedsheridan8725 2 місяці тому

For the squeeze theorem part, all you need on the lower bound is 0. X!:x^x is greater than 0 as all factors are positive.

@ytyayagamer2303 13 днів тому

Will extend to 1 as x(x-1)/xx =1(1-1/x)and when extend to infinity it equal 1(1-0)=1 this for x Permutation 2/x power 2 so if we made extend to infinity it will give same answer

@herodotomello 4 місяці тому

Man, you're great! Greetings from Brazil

@dennisestenson7820 2 місяці тому

Before watching it, it seems obvious that the limit is 0. The numerator is the product of all the numbers between 1 and x. The denominator is the product of x with itself x times. The denominator obviously grows much faster as x grows. So the limit is 0 as x approaches infinity. After watching, I see the squeeze theorem would be the formal way to show the same thing.

@indescribablecardinal6571 5 місяців тому

I like sandwich theorem a lot, I liked this way. When I saw the thumbnail I tried to solve it, and went to the path of separing the product expansions in corresponding ratios (x/x, (x-1)/x, (x-2)/x, ... , 2/x, 1/x) and demonstrate that the limit of all that factors are either 1 or 0, and a solely 0 as factor is enough to make the product equal to 0 (unless if there is an infinite or an undefined/indetermined factor, but this is not the case). I went on the brute force 😅

@SidneiMV 5 місяців тому

Awesome approach! 😎

@gilmartrevisan Місяць тому

Great job, Professor

@jakubfrei3757 5 місяців тому

Im always waiting for a moment when this guy start rapping and its not comming, which makes me feeel a little bit awkward, but i really love the way you explain things, good job !

@PrimeNewtons 5 місяців тому

Someday!

@jakubfrei3757 5 місяців тому

@@PrimeNewtons Im glad You're not taking it in bad way friend!

@editvega803 Місяць тому

Beautiful! 🤩. Thx!

@kaushikchakraborty1692 5 місяців тому

Excellent explanation. Thank you.

@manuelgarrido5602 5 місяців тому

A pleasure to watch! Tx u!

@wolfix20021 5 місяців тому

Thank you sir!

@conceptswithpramod 5 місяців тому

Thank you making a pretty straight forward question so complicated. x factorial = x.x-1……3.2.1 Now take out x from each term x factorial = x^x(1.1-(1/x).1-(2/x)…2/x.1/x) Hence numerator and denominator in question cancel out and limit is equal to 0