does your Calculus teacher give you limits this hard?!?

[@CamiKite]

Always satisfying when you see a Riemann sum appear

[@mattcarnevali]

I’ve never considered anything but an integer as the option for the nth-root. The n^2 root is wild

[@wesleydeng71]

Really?😅

[@shirou9790]

ah yes, n^2, famously not an integer

[@leif1075]

What do you mean..n is an integer per the problem

[@postbodzapism]

​@@shirou9790It is a polynomial non-integer.

[@canyoupoop]

We do a^x all the time x∈ℝ. His x√ notation just make it weird

[@eduardsluzhevsky6265]

Thanks!

[@theproofessayist8441]

thank you for going through this problem - I would be traumatized if this limit problem appeared on an exam.

[@goodplacetostop2973]

11:15

[@pragyanpranay3681]

dude is still on it. I love you bro ❤

[@AndyBaiduc-iloveu]

Weren't you the guy that gave homework problems?

[@Aditya_196]

​@@AndyBaiduc-iloveuyes he is

[@АртурАнищенко-ю2п]

saw this cute one on IMC :) it also can be solved easily by Stolz' theorem, though it's not as elegant

[@thomasidzikowski1520]

Super enjoyable as always.

[@dentonyoung4314]

That was a great mental review of algebraic tricks and limits.

[@leif1075]

This method is too contrived and not fair isnt it? And it'snotna real intrgral necause m only encompasses the integers and ofr an integral n must take all real values on theinterbalm.no one is going to think to write n/n times m..seriously why would they..it's not smart and not logical and just random and I'm tired of that..

[@2070user]

@@leif1075If you are tired of it, just stop watching and spamming these comments. It saves everyone's time, have a good day.

[@AndyBaiduc-iloveu]

Plz just get out of here , practice a lot and make a great foundation of math of your level , then go to the next level , one by one . Be patient and you'll know that it's not so random.

[@darksecret965]

The interbalm 😂😂

[@solcarzemog5232]

Freaking amazing!

[@Yougottacryforthis]

Cool stuff like that sum turning into an integral

[@catburner1896]

Looking initially, the answer is zero because the sqrt can be rewritten as mult as x->n; x^x/n^2 And because n>=x for all x, x^x/n^2 as n->inf is zero, And sqrt(inf) is going to be big, but as it is in the dominator it will act as a zero. This gives us 0*0=0

[@byteeater7662]

I really don't want it to sound as any kind of harsh criticism, but sometimes Michael goes for a grossly overcomplicated method, and this is one of those. Well, happens to most of us, including myself probably more often than I even realize, because prevalently likely unbeknownst. Here's my take. What remains after applying the ln and pushing it under the lim (because continuous) is a difference of fractions with denominators n² and 2, so it can be transformed into one with common denominator 2n² which lends itself well to the Stolz theorem (arguably simpler than de l'Hospital's rule, being its discrete version; and thereby you avoid integration and other complex stuff Michael does). The final unobvious touch is using the known fact that (1+1/n)ⁿ (which appears under ln) converges to e.

[@staswisniewski4101]

There's also nice fact, that if a_{n + 1} / a_n has limit g, that \sqrt[n]{a_n+1} also has a limit Could we somehow use this?

[@chrissquarefan86]

This is exactly the kind of limit I enjoyed in my high school years

[@sardine_man]

To solve the last integral we can also invoke the gamma function relatively easily

[@Bangaudaala]

Can someone explain to me why at the end we dont get an undefined case of infinity*0? We have (t*e2t)/2 from 0 to -infinity We get 0*1/2 - (-infinity* e^-infinity/2) which is 0+infinity*0 Doesnt make sense to me why were left only with -1/4 Edit: Ok I did check the limit with L'hospital but why didn't he write that, he wrote it for the other L'hospital case

[@MrEddrool]

L'Hopital confirms what the professor knows instinctively, which is that the exponential term dominates that limit and makes it zero. That's why he skipped talking about it.

[@harrymattah418]

Formally, I would rather label (say f(n)) the expression we are searching the limit L thereof, then consider ln(f(n)) and investigating for a limit l thereof. If such a limit is found, then properly justifying that lim ln(f(n)= ln (lim f(n)), to conclude that L=exp(l)

[@pavlopanasiuk7297]

Pretty cool idea, was able to derive myself. I wonder* if there's another method in plain sight, to me this seems the only way to navigate..

[@forcelifeforce]

* wonder "Wander" means to "walk or move in a leisurely, casual, or aimless way."

[@leif1075]

What?? wWHAT The muktiplying m by n/n is COBYRUVED and I don't see why anyone would think of it and therefore shouldn't be alowed or done. Don't you agree? Seriously it's just random

[@pavlopanasiuk7297]

​@@leif1075 "shouldn't be allowed or done" because you failed to find reason? I'll ignore that. Reason has been the sum seemed a lot like Riemann sum, it just needed some derivation

[@NullClass]

You can also solve it using Cesaro-Stolz. Also, not 100% sure, but I think a simple squeeze on the series that comes up after applying the log is enough. You can use the fact that it is monotonic to estimate it with an integral. Will try it on paper later

[@FLCoeur]

Glaisher constant after it

[@postbodzapism]

I wonder if Stirling's approximation gives an alternative method.

[@AndyBaiduc-iloveu]

At first glance , I don't think Stirling approximation is useful here . I'm not sure though

[@canyoupoop]

Using factorials makes it worse

[@postbodzapism]

Probably an allusion to multiplicative calculus is appropriate.

[@Bodyknock]

7:57 I wonder why he didn’t just do integration by parts on xlnx directly? If you differentiate lnx you get 1/x and antidifferentiate x it’s not that hard from there. (You just need a use of L’Hopital’s Rule in the solution and you end up with the same solution as in the video.)

[@xinpingdonohoe3978]

It may be because he wanted to show off how we like to differentiate the polynomials, as after finite iterations they become 0, and we like to integrate exponentials, as they only scale whilst keeping the same form. This wouldn't be what we'd do when we differentiate ln(x) and integrate x.

[@pierrot31511]

I don't like writing lim when we still don't know if it exists.

[@juanmanuelcaceres4142]

I think there is a mistake when you put exp(2t) instead of exp(t). Great video.

[@SpandanJawkhedkar]

There is no mistake. x•dx = (e^(t))•(e^(t)dt) = e^(2t)dt

[@Neil.Menezes]

No, he knows my limits.

[@wannabeactuary01]

Excellent

[@yoav613]

Nice!

[@diniaadil6154]

Using an integral comparison i quickly got exp(-1/4)

[@leif1075]

Really what test? Whybintegrals though when we are not dealing with integralsals? What made you think of it..rhe fact we have an infinite product? That still by itself wouldn't explain why you'd think of integrals and would youbagree most ppl wpuldnt..think ofnit unless you'd seen it before? And is it because younthoight of ln x beung represented as an infinite series that led you to think of it?

[@diniaadil6154]

@@leif1075 I just got a stroke reading your comment. Basically yeah : approximate series of n*ln(n) using integrals gives you n²ln(n)/2 - n²/4.

[@ANTONIOMARTINEZ-zz4sp]

It's just fascinating... 😏

[@alexchan4226]

0

[@nuranichandra2177]

How are we sure that the logarithm of L exists and is defined?

[@bertrameiriksson9274]

L is clearly positive.

[@dominofan238]

His calculations actually show that the limit exists in the first place. Up until 6:51 it's just rewriting ln(original expression) without taking any limits, so it would have been cleaner to simply not write lim yet. Then, since the limits of the two summands both exist (one is -1/4, the other is 0) the limit of ln(original expression) exists and is -1/4 and hence lim (original expression) exists and is e^(-1/4). The last step makes use of the continuity of the exponential function.

[@leif1075]

MICHAEL CAN YOU PLEASE RESPOND..NONPNE is going to think ot write m times n/n NONPNE SP WHY DO IT..And it's not an integral sonce ypu dont hwve all real values between 1 and n

[@hedgedog-gh7zs]

I''m assuming you mean the step at around 6 minutes, when he converts the limit of the sum into the integral. It's just using the definition of the Riemann integral - it doesn't require all real values between 1 and n.

[@Fred-yq3fs]

The idea when you have a sum is to try to convert it to an integral. You do this by "bending the sum to your will". To make a Riemann sum, you need points between bounded values, say 0 and 1, so you'd like Ln(smth between 0 and 1), so you try m/n, and lo and behold you have enough n outside the sum to also bring one in to form m/n. And voila: you start to recognize the integral for xLn(x). The "magic" introduction of m/n is part of the craft: you've got to identify a known pattern, and get there using all the tricks you know.

[@Fred-yq3fs]

Honestly, this part of "craft" requires practice. It won't come to you for free. You have to try your hand many times, get inspiration from gnarly problems, build your pattern recognition, develop your tricks... For most students, abilities have to be earned, and hard.

[@martincohen8991]

Just do sum m \ln(m) ~ int_1^n x ln(x) dx = n^2 ln(n)/2-n^2/4