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.
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.
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)
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 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
@@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.
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.
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.
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
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.
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!
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!
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!
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.
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.
@@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.
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.
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.
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..?
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.
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!
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! 🙂
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.
You are absolutely right in case of natural numbers whereas we take limit of function over real numbers. According to your definition of factorial, it is not even defined on real numbers so that leads everything that you have done to be actually completely no sense.
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 !
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.
Nice presentation, and clear explanation! I was pretty sure 'by inspection' (sorta guess) that the limit was 0, but couldn't prove it. Now I can! Wheee! Thanks.
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
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.
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
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
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. :-)
I don't know why I'm watching tricky limit problem at 3AM while it's been years ago I was graduated. Surprising result though, I didn't expect that. But thinking about it, makes sense actually, the denominator grows much faster. Anyway, you just got a new subscriber, you're an awesome teacher.
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.
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.
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?
