A Brilliant Limit 

Noticing that the limit is actually an integral is beautiful. I've never seen that done before. Not having to use any advanced theorems, just basic calculus methods is a pure joy.

I cry a little inside every time I see him use a pen that isn’t Black or Red ...

Cauchy's theorem on limits works fine on this one. Take a_n = n!/(n^n), it's a sequence of positive numbers. The theorem says: Let a_n be a sequence of positive members. Limit of a_(n+1) / a_n exists. Then lim (a_n)^(1/n) = lim a_(n+1) / a_n. These conditions are met, so we can use it. After cancelling those factorials and n+1 in powers the expression leads in limit to 1/e. You even do not need integral aparatus for it.

We can also use the Stirling's approximation. When n tends to infinity, n! ~ sqrt(2πn)*(n/e)^n Which means (n!/(n^n))^(1/n) ~ (2πn)^(1/2n)*(1/e) ~(2π)^(1/2n)*n^(1/2n)*(1/e) So the limit is 1*e^0*1/e = 1/e

n-factOREO

So... series gave birth to limits which gave birth to derivatives and integrals.. Now we have this limit problem that can be expressed as a series problem which is, in fact an integral problem... Maths is like a thriller movie: whatever the starting situation is, you never know which character is the turning point in the end. I love thrillers.

The instant I saw this problem, I figured it would involve Euler's number. It even looks similar to the definition of *e.*

Nice job! To integrate y=ln(x) from 0 to 1, turn the function on its head. Just integrate x=e^y from minus infinity to zero, which comes to 1, and take the negative of that, which is -1.

Waking up and watching a math video from the best RU-vidr while having breakfast, this feels good.

Great approach via the integral, I ended with the same result using Stirling's approximation; only to have a rogue scalar of \sqrt{2 \pi} alongside e^{-1}.

Beautiful! A really nicely presented example of the beauty of mathematics. Incidentally, 1/e is the unemployment rate in the limit when a large number of employers randomly chooses to hire a worker from among the same large number of workers. Euler's constant shows up everywhere, like a mathematical Zelig.

I actually cried out "holy shit" when you converted it to an integral. That was real eureka moment for me!

You can do the integral pretty quickly, you can realize that geometrically it is the exact same as the area from -infinity to 0 on y=e^x , which of course is very quick to do as its just e^0 - e^(-infinity) = 1 - 0 = 1, and then since its below the y axis its negative. Maybe its not rigorous though?

谢谢老师。我越看越惊艳！ Although my math knowledge is limited (due to my own laziness, nowadays there are no excuses to learn), I really enjoy your videos and actually learn a lot... And of course get amazed almost everytime. Math is extremely beautiful and rewarding.

I've been using this approach since an year but today's the first time when I truly understood it. Thank you so much.♥️🙏🏻😊

I love how happy this guy is just to do the maths and share it. It makes me smile. 😀

brilliant question and solution

My first reaction was to use Stirling's Approximation for n!. Returns 1/e pretty quickly. (n!/n^n)^(1/n) ~ ((n/e)^n * sqrt(2*n*pi)/n^n)^(1/n) (n!/n^n)^(1/n) ~ (1/e) * sqrt(2*n*pi)^(1/n) (n!/n^n)^(1/n) ~ 1/e

Very nice use of recognizing the series was equivalent to an integral. I had started the problem using the gamma function, and restricting to integral values, but your solution was much more elegant. Nice job.

This man is an incredible mathematician. He makes me go "oh fuck how did I miss that, it seems so obvious now".

Here is a much simpler solution: It is well-known that if for a sequence a_n of positive numbers the limit of a_{n+1}/a_n exists then so does the limit of the n-th root of a_n, and both are equal. For the obvious choice of a_n, here we have the ratio of interest equal to (n/n+1)^n which converges to 1 over e. The end. And the result referred to above is more elementary than information on Riemann integral.

Mind. Blown. That was amazing.

What an absolutely brilliant solution! This is one of the coolest limits I've ever seen, thank you so much for showing this to the world!

This is one of the most brilliant and beautiful solutions I saw you doing.

Great video! I used Stirling’s formula after doing the logarithm of the limit instead, that way you can transform ln(n!) to n*ln(n)-n.

Holy cow. How do you relate that to a definite integral from 0 to 1?! That's pure genius omg.

I really enjoy your videos, they help me solve Calc problems a different way. Continue making more!

Love how excited he is about solving and explaining math exercises, keep it going

I find it really fun that you're showing us all of these examples that 0^0 isn't always equal to 1. That it depends upon the function that you're taking the limit of that takes a 0^0 form at the limiting value. Also find it really fun that virtually every single problem you come across that can be solved in multiple ways, your method almost invariably (pun intended) involves an integral at some point. You love your integrals, don't you? ;)

This video made me look at math from a whole new perspective! What a beautiful question

The moment you said you couldn't differentiate n! I was like: actually there is... then you mentioned the Gamma-function and I was happy

Two minor suggestions for streamlining the presentation: First, I would have just written n! as 1*2*3...n rather than n*(n-1)*...*3*2*1. This is easily accepted up front and avoids having to reverse order later on. The other thing is just before your discussion of the sum being an integral, I would collapse the sum of logarithms with summation notation, e.g. sum(i=1..n) ln(i/n) * 1/n. The sum->integral discussion then becomes easy to follow with the equation in this form. It's easy at that point to just call x=i/n: the limit of the sum from i=1..n then directly becomes the integral from 0 to 1, and the 1/n is directly the dx for closing your integral.

This math problem is brilliant!

im studying very advanced math in my university and i had my doubts watching the whole video but man i really loved it. i wish my professors would have your enthusiasm. it really is a beautiful limit! and i think students should be shown these kinds of examples to see the true power of infinitesimal calculus.

Taking AP BC Calculus has given me a whole new side of math to love, we haven’t gotten to integrals yet but this limit made me excited for it

I have vague memories of going from infinite product to an integral back in Highschool. Watching your video gave me a wonderful aha moment as the recollection hit me! Thank you!

what a creative solution. I love it!

You pretty much explained how Stirling’s approximation works. Excellent video

Amazing solution! Another much easier way to work out this limit is to use Stirling’s formula, which states that n! = (n/e)^n • sqrt(2•pi•n) • (1+epsilon(n)), where epsilon(n) goes to zero when n goes to infinity. Using this formula, you almost immediately get the answer 1/e.

Hai fatto capire benissimo la tua esperienza esilarante. Complimenti per il video!

This same question came in our Jee Advanced Mock tests last day and today I came across it on youtube..😂 What a coincidence

This question was once asked in IITJEE When they have subjective paper I had solved this

You are excellent! You always bring a new level of creativity to solve the problems.

if you want to calculate integrate of lnx between the born of integration you can also use the bijection which is exp(x) in order to make it more easily to respond

In India in jee adv examination for IIT entrance this type of q is asked Very similar q was in jee adv 2018 and jee adv 2017 and many earlier also. We study this topic as "limit as a sum" .

One of the finest questions I have seen u do. Keep up👍

That integral trick gave me Goosebumps!! Brilliant

Woooaaah, that was crazy dude good job! I think it's too crazy to show my calc class but still good job!

beautiful. just beautiful. very elegant. reduces to a right Riemann sum after all.

je suis vraiment étonné du nombre de gens qui regarde la vidéo alors que l'on vie dans un monde ou les maths sont de plus en plus mis en retrait (en france en tout cas). Je trouve cela vraiment bien et merci pour la résolution.

maths is beautiful, by far the coolest limit i've seen

Makes me feel humble. i would never have got that in a gazillion years.

I wouldn't have let you pet me any longer if you had said IN FRONT OF ME, that I was not as fluffy as the other one, aswell..

A really good teacher, I will enjoy your videos! Well done

I loved it. Thanks for such an layman level explanation! :D

Brillant indeed! I wonder why the anti derivative of ln(x) is not one we are asked to learn in school, even though it's pretty easy to remember.

Solved this while preparing for IITJEE

I loved watching this - /finally/ subscribed to your channel!

My teacher give us the same limit six years ago and I used Riemann integral to solve it,this video bring me some great memories.

10:00 and elsewhere Anyone else notice that, when writing "f", he does the horizontal stroke first and then the ∫ ?

I used the Stirling formula :D maybe it is cheating....

I remember this technique from the secretary problem! It's so satisfying turning infinite sums into integrals!

How is this guy so entertaining. It's actually insane.

Am I the only one who solved this using stirling's approximation for n factorial? It makes the solution super easy: Sqrt(2pi*n)*(n/e)^n < n! < (e^(1/12n))*sqrt(2pi*n)*(n/e)^n If you replace the n! In expression in the limit with these two expressions the n^n bit cancels out leaving some expression taken to the power of 1/n which clearly converges to 1 as n goes to infinity, multiplied by a constant 1/e. Placing the expression in the limit between the two in a chain of inequlities and taking the limit of the upper bound and the limit of the lower bound, the result clearly goes to 1/e and by the sandwich rule we get that the limit we are evaluating must go to the same limit as well. This is so much quicker than what was shown here. However, his solution is still a neat one and I get why he would prefer to show it rather than the quick and easy route I took, which may seem like trickery to someone unfamiliar with this inequality.

Since I wondered about this myself: the reason that you cannot simply take any base b for the initial logarithm operation and then find a solution of 1/b is that the processes are only the same until the integration step. You can pull out the limit for many values of b (where logb(x) is continuous with x>0). You can always pull the exponent of the argument into a coefficient; this is a basic property of logarithms. Then another basic property is used to turn the problem into a series. However, at this point, it is demonstrated that the series limit is equal to the integral from 0 to 1 of ln(x), which is equal to -1. It is still correct to say that that limit finds the integral of logb(x) from 0 to 1 of any b (for which the function is continuous from 0 to 1), but the answer will not be -1 if we use another base b! Recall what we are finding: this limit (which is also an integral) is equal to ln(L), or log2(L), or logb(L). Which integral we find depends on which logarithmic base we used first; log2(L) does NOT equal the integral from 0 to 1 of ln(x), but of log2(x), which is a different number (-1 is only so clean because e is special). Then, we should raise our base b to the value of this integral, to solve for L. That number, if you multiply it by e, gives you 1; that means it is 1/e. This is the part where e comes in whether you like it or not! Since our original limit only had one answer, we can use different bases to get it, but the combination of integrating those bases and then raising them to the power of that integral always gives us 1/e. You can try this with base 10: the integral from 0 to 1 of log10(x) dx is -0.4342944.... Raising 10 to this power gives us 1/e, which you can prove by multiplying by e to receive a number very close to 1.

I clicked this video because i was bored and expected it to be clickbait, but boy, was this a brilliant limit. Had me captured for the full 15 minutes.

I'm 17 and solved this in 2 minutes through summation and integration..saw that you used same method ..Felt good. Did many problems like these will i was 16 for sharpening calculus..

I'm on that website as well!!

the tittle must be "most emotional moments of 21st century" or smt...

Putting equivalent to integration shows really understanding differentiation and integration of limiting nature . Good and Brilliant!

very intuitive and simple approach, thank you! 😊

although I am bad at English，but I can understand your analysis of this question

This is truely beautiful

You are a genius mathematician I like your style Love from India

Very clever solution, love the integral insight!

Sterling's approximation of n! would also do it.

"Imagine you're just petting her and do math"

Since the integral from 0 to 1 of ln(x) is the area unther the x line you know that that is the area of the function e^x from - infiniti to 0 since a funtion and its inverse are simetric along the line y=x, and the integral from e^x from - infinity to zero is simply e^0 - e^(- infinity) = 1, since the integral of the ln(x) is below the x axis the integral is -1

Graphing is a powerful method in reasoning, the solution to problems in calculus. He had to evaluate the integral in the end ( integration by parts ) to get xlnx -x + C as the answer. Evaluating the upper limit of integration: 1*(ln(1) -1 ) + C which is 1*( 0 - 1) + C = -1 + C. Now the lower limit of integration: 0 * (ln(0) -0 ) + C = 0 + C. Subtracting from the lower limit of integration the upper limit of integration: ( -1 + C ) - (0 + C) = -1 .

looking at it and thinking about it for 15 seconds, i feel like it should go to 0, but in my head i know it’s somehow gonna go to e edit: well, close enough

Could you solve it with gamma function please?

Yey. The right problem I'm looking for. Thanks!

That's a really creative way of solving this problem👍 brilliant!

OH MY GOSH I MADE THIS LIMIT YWATERDAY BEFORE I WENT TO BED

I knew it had something to do with e, I felt it coming....

I teach AP Calc and have never seen your videos before. This is awesome-- I just subscribed :)

An alternative, and perhaps easier solution would just be to use a conclution of the stolz-scesaro theorm ("L-Hopital rule for sequances") which is that for any sequance an, if the limit of (a(n+1)/an) converges, then the limit of n'th root of (an) converges into the same value. So all we have to do is just calculate the (a(n+1)/an) limit (which is super easy) and we are done. (This may sound complecated for those who aren't familier with this theorm but it's a very very useful one for calculating complecated limits and series Although I have to admit that bprp's solution is freaking awsome!).

No need to integrate by parts. You can change from a dx integral to a dy integral, and it becomes straightforward. integral(ln(x) dx, from 0 to 1) = - integral(exp(y) dy, from -infinity to 0).

I correctly guessed the answer by doing an Excel spreadsheet up to n = 143.

We can demonstrate it using "D'Alambert's quotient limit implies Cauchy's radication limit", but we lost a brillant method and application of Riemann's sums. Thank you for an excellent video. King regards from Argentina

This is why I love math, it's beautiful.

Crazy indeed

taking natural log on both sides and interchaging log an limit on right hand side is not permitted unless the limit exists..by composite functions therorem.

I am watching because I saw factorial inside limiting function. Thanks for bringing up cool math problems.

I absolutely love your videos!🔥🔥🔥

他英语说得很好！

how come the number "e" appears so often in limits? can someone explain this to me?

Brilliant! :) Thanks for this. I enjoyed it!

Brilliant solution ! Loved it.