A COUNTER-INTUITIVE CALCULUS LIMIT 

Casually doing math while holding a thermal detonator.

It's clever for you to use the blue pen for "I did this in a different video" because it makes me think your whiteboard has hyperlinks. When I go watch that video and come back will that part turn purple?

The "please pause the video and first try for yourself" is always so motivating :) I was lying in my bed and this motivated me to get up and try it :)

It helps when you keep in mind that the arithmetic rules for limits of sequences only apply if all sequences used actually converge :)

That it's negative this makes sense considering that limit definition of e approaches e from the left of e.

I’d say 0. “en” means “one” in Danish, so it is 1 - 1.

For the "intuitive" explanation: The 2nd part, x*e, is like a typical linear function: y=k*x, with the constant being "e". This of course grows linearly. The 1st part, x*(1+1/x)^x, will eventually reach a similar value, the latter part eventually becoming "e", so this term will also become a linear equation essentially, x*e. But it takes a while for it to grow and get closer to the value that is "e". Thus x*(1+1/x)^x is "lagging behind" the other function. After a while it will stabilize though when becoming more linear. The difference between the two functions caused by that lag will thus be e/2 (the minus sign in the limit depends on which of the functions we put first). If you want to see this yourself, plot the two graphs: y(x) = x*e g(x) = x*(1+1/x)^x And measure the height difference for some relatively high value of x. Should be about e/2 = 1.36.

Weird things can happen with infinity. This was one of my favorite videos so far!

That was some fast writing at 6:44 and 6:48! 😁

I was having a crisis thinking Q1 might not be 0 somehow lol

Using ln(1+1/n)=1/n-1/2n²+o(1/n²) and e^(1/n)=1+1/n+o(1/n) you can solve that easily and quickly without using l'hôpital rule and without these weird derivatives. Taylor series are so useful to find limits... You make this looks really hard when it really isn't.

Yesterday the beard ... today the plaid flannel shirt ... tomorrow - full lumberjack? edit: re-watching after three weeks (forgot method - whoops) I felt the desire to add: time to remember it was "x", not "n": about 27 seconds - been there!

Believe it or not, I did it in my head, using limited developments of ln(1+x) as x goes to 0 and of exp(x). Did it like (1+1/n)^n - e = exp(n ln(1+1/n) ) - e ~ exp(n(1/n - 1/2n^2)) - e = exp(1 - 1/2n) - e = e(exp(-1/2n) - 1) ~ -e/2n And then multiplied by n it leads to -e/2.

You can also use the Taylor expansion twice, with ln (because (1+1/n)^n = e^nln(1+1/n)) And again with exp (you factorizes by ne : ne*(e^-1/2n - 1) and the Taylor expansion of exp(-1/2n) gives us 1-1/2n so finally ne*(-1/2n) which is equal to - e/2 I find this quicker ;)

That moment when you forget a minus sign

In the process of looking more into your video about generating x and y such that x^y=y^x, I found that (1+1/x)^x and (1+1/x)^(x+1) generates such a pair (directly derived from your method using t). Since these pairs always bracket e, and are almost equidistant from e when they are close to e, a better estimate for e is (1+1/n)^(n+1/2). If you substitute this better estimate for e in your expression and then factor out (1+1/n)^n (and call that e like you did in this video) and then find the limit of the remaining factor, also using the fact that for large n, sqrt(n(n+1)) is about n + 1/2, that limit is quickly found to be -1/2, giving the correct answer -e/2 (not by such a rigorous process, but you might be able to justify it better than I could). No LH rule or derivatives involved, just algebra. Edit: I went through the LH rule procedure for the limit of ((1+1/x)^(x+1/2)*x - e*x) and unless I made an algebra error, this limit is 0. The polynomials in numerator and denominator are one degree higher, but the numerator still cancels out to 1. [Just checked it on Wolfram Alpha, the limit is 0.]

One of the best math channels out there - deserves more subscribers! Keep up the good work!

The answer makes perfect sense since that would be the accumulation of errors between the limit and the value e. I expected it on the value becoming negative but arriving on exactly -e/2 was pretty amazing

the lady at the beginning is absolutely gorgeous

I love this channel! It's like the BPRP of RU-vid!

You are not right on 4:20 It does matter, becuase "n" is natural and "x" is real. You can't just replace first by second. For example, sin(2*Pi*n) converges to 0, when n=+infinity, but sin(2*Pi*x) diverges when x=+infinity. Would be interesting to watch video, where you explain, when we can do such substitutions, and when we can't.

Another approach is to do the binomial expansion of (1+1/n)^n and then subtract the series expansion of e. Multiply the difference by n and the answer falls out. No l’hopitals needed.

70% of the time i was looking at his beard

Without using complicate tricks: n(1+1/n)^n = n exp(n(1/n -1/2n^2 +o(1/n^2))=nexp(1-1/2n+o(1/n))=ne(1-1/2n+o(1/n))=ne-e/2+o(1) So the limit -e/2 is straightforward

Stop blowing my mind!! Actually, keep going. Fascinating videos that bring back some knowledge I haven't used in a long time.

let's solve this question with very unnecessary complicated, set a complex function f(z)=(1+1/z)^z. now we have f(n)n-en for n to infinity and Im(n)=0, we can calculate Laurent series of f(z)(i won't write here the calculations) to get f(z)=e-e/(2z)+o(1/z^2) hence Lim f(n)n-en=(e-e/(2n)+o(1/n^2))n-en=ne-e/2+o(1/n)-en=-e/2+o(1/n)=-e/2

I've never seen you use the Landau notations ( a_n = o( b_n ), a_n ~ b_n, a_n = O( b_n ) ) which make this exercise pretty easy without using derivatives ! Is this not in use in the US ? Because we use it a lot in the French system, and I must say, they're very useful !

is it really acceptable to just throw l'hopital's rule around like this before checking if the condition for it even apply?

4:30am. Good morning, time for BlackpenRedpen

Hi BPRP, can you do a video on the integral of cos(x).sqrt(1-cos(x))? Thanks and I love your videos by the way!

I posted this comment at flammable maths as well. This was pretty easy if you do a binomial expansion on (1+1/x)^x = 1 + 1 + (1/2!)(1-1/x) + 1/3!(1-1/x)(1-2/x) .... 1/x!(1-1/x)...(1-(x-1)/x) subtract e = 1 + 1 + 1/2! + 1/3! ... multiply through by x. This will leave a collection of constant terms and a collection of 1/x^n terms. You can disregard the 1/x^n terms as they go away as x get to be large. Summing the constant terms you get -1/2(1+1+1/2! + 1/3! ...) = -1/2 e.

This guy's living the dream: beautiful wife, genius son, wears stylish clothes, makes math videos for a living.

Nice video, cool limit I'll try to remember to be careful when n is doing "multiple tasks" in my limit. I feel like it's a really easy thing to misunderstand or accidentally interpret incorrectly.

You make limits interesting!

I would have used ln(1+x)=x-x^2/2+O(x^3) to work out an approximation of ln(1+1/n)

or through a numerical approach you can put n as equal to 1.0E9 (or preferably more) and you get the answer as -1.35904 ~ - e/2

yes of course! Limit calculation form of 0×∞ should be treated carefully!

Tried running a code to check this limit, and somehow the limits start blowing up after 10^9 it seems... Does Python not support the amount of precision required for the limit to approach -e/2?

Loved this video! Made me think. I immediately went into desmos and plugged in all of the functions and discovered the -e/2, thought about it, then came back to finish the video.

3:03 it's called desamol or desimait.. (i am not sure) you can find it serching for "the opposite of infinity"

You and Peyam are the best

"為什麼這個是，欸" lmao I know you can speek chinese but have not evidence of it enough, and this video just bring me laugh. I like your video, and hope you can make more and bring us fun.

Brilliant! Your enthusiasm is infectious!

This actually gives information about the convergence rate of (1+1/x)^x to e

THANK YOU SO MUCH!!!!! I das hours on a similar question...all because a little mistake I didnt notice myself. Thank for the video

I like that you post your mistakes as bloopers either at the beginning or end of the video. It shows us that even the great blackpenredpen is still ultimately human and makes mistakes like the rest of us! :D

Wasn't this incorrect when you used L'Hospital's rule to solve lim( ((1 + 1/x)^x - e)/(1/x)) ? One can use L'Hospital's rule for solving uncertainties of {0/0} and {inf/inf}, like, lim( f(x) / g(x) ) = lim( f'(x)/g'(x) ). But it only works if g'(x) != 0 when x → inf (or zero). In your case, g'(x) was -1/x^2 and it is =0 when x→inf. I think using L'Hospital's rule was incorrect in this case.

"derivative of (1+1/x)^x, it will be here soon, derivative of ln(1+1/x), it will be here soon"

I would have thought it was 0*n which is... Problematic...

2:44 can you really say that the limit is "e - e", since the limit operator does not distribute over the individual terms unless each component converges, but "n" itself goes to infinity so we cannot say that converges?

In response to Andrea Parma, there is another way to get this. If one uses Sterling formula for the factorial, then the same results falls out without using L'Hospital's rule. n! ~ sqrt(2 pi n) * (n/e)^n The "e" falls out from rations of (n/e)^n for n and n+1 and the -1/2 from ratios of sqrt(2 pi n) and sqrt(2 pi (n+1) ). It is a round about way that uses integral results (ie Sterling) rather than derivatives (L'Hospital). PS blackpenredpen: I am growing a beard if I can get a nice chick like your announcer....

great example on being careful with limits! where is this from? I want more!

The comment @4:10 is some what misleading. If limit as (real) x-> ∞ of F(x) = L, then limit as (integer) n -> ∞ of F(n) = L. However, limit as (real) x-> ∞ of F(x) may not exist but limit as (integer) n -> ∞ of F(n) may exist. Example: F(x) = sin(pi*x)

Wouldn’t this mean that the value of the limit would be the same no matter what finite number you used in place of e!?

the lovely doraemon sequence . love you man

If you factor out the n and divide it to the other side you get the error for the exponential approximation of e... in other words: Error = e/2n And approaches zero as n goes to infinity. Which implies the approximation is getting better. Just my 2 cents... haha

I have only one question: who is the girl?

Nice limit. I've got the same result using Taylor's formula.

Or, at 7:18, instead of using l'Hopital's rule a second time, you can just bring the -1/x^2 in the denominator into the numerator as -x^2.

Interesting problem, thanks. Though am wondering: why rename n to x instead of just writing d/dn?

I have never seen a limit where using l'hopitals rule twice leads to the result.

As x approaches infinity, why can we ignore the +2x on the bottom of the fraction?

This limit is a curious example of the subtlety with which one works with the infinite potential and the actual infinity.

Another way you could do it is take (1 + 1/n) = e^(1/n) - 1/(2n^2) + O(1/n^3) and go from there.

11:37 You desappeared at the last frame. O_0

You can also take the logarithm of the expression to begin with. That will make the derivative simpler.

OMG So interesting

@blackpenredpen, can this be interpreted as if (1+1/n)^n approached e as fast as n grows to infinity? Or, equivalently, their difference goes to zero as fast as 1/n does?

I thought you were going to set x=1/n. That would put just x in the denominator, you differentiate it and get 1, and you just have to differentiate (1+x)^(1/x) at 0.

your addicted to L'h rule like seriously your using it in every lim video

Visually if we graphed the 2 functions F(x)=(1+1/x)^x and G(x)=e.x, at the infinity G would have a 1.359140914 lead

I wish we had math like this at school instead of boring lectures

If you want to differentiate such things, better make the substitution t = 1/x in the limit. When x grows to infinity, t approaches zero.

Why not just put L(n)=ln((1+1/n)^n) = n ln(1+1/n) = n(1/n - 1/2n^2 + 1/3n^3 - ...) = 1 - 1/2n + 1/3n^2 - ... Then exp(L)=exp(1)*exp(-1/2n+1/3n^2-... = e*(1-1/2n+1/3n^2-...), n exp L = e*n - e/2 + O(1/n) which is basically what we need

Another easy method using taylors theorem,let x=1/n so limit becomes lim x tends to 0 of ((1+x)^(1/x)-e)/x now use Taylor exapnsion of (1+x)^1/x which is e(1-x/2+11x^2/24+O(x^3)) and solve it very easily to get -e/2

the last 4 steps can be simplified if u replace ln（1+ 1/x）with 1/x. 1/x-1/1+x can be cancelled by the denominator 1/x².

ooh, a rare outtake. I totally never have moments like that.

Sir too algebra, that's not the issue but ina exam hall when the timer is going going gong, there has probality to be wrong So I use series expansion.it's really quicky way to solve..

Sooooooo amazing !!!!!!!!!! I think its not zero beacaus when we have e.n its biger then( (1+1/n)^n )*n Beacaus that is already(e) but the first one we have to go to infinity to get (e) so its like a race when two people is the same speed but one of them begin to run first !!!!!!!!!!!!!!!

1:39 Doraemon?

Haha lovely intro

Hlo sir I am big fan of yours from Nepal could you please solve the important question from undergraduate scholarship of Japan that is mext Scholarship program some questions are feels hard

Keep up the good work, enjoy your Channel!

Why can't LH's rule be used on the first 2 limits? inf-inf and inf*0 are both indeterminate forms.

Is it possible to do without L'H rule? Cuz my teacher restricts me using it

Where do you get such type of problems? Please, can you suggest me?

Checking the result: Put 1,000,000 or even 100,000 for n in the original lim and calculate. it gives -e/2 in a very high accuracy.

If it does not matter if it is x or n then why do you change it? Just differentiate in respect to n.

4:55 can you apply l’hopital here even though (1+1/n)^n is an indeterminate form if you plug in infinity? Or can you just use the limit and say it’s e…

If you want to know the derivative of (1+1/x)^x you can take the derivative of e^(xln(1+1/1x)) which is the same thing that he got on the screen

I have a question . please help me spotting my mistake : f(x) = ( tan( πcos²(x) ) / sin( 2πsin(x) ) ) task is to find limit as x->0 what i did was multiply and divide numerator by π * cos^2(x) and as α->0 tanα/α =1 , and doing same with denominator sin(α)/α form πcos²(x)/2πsin(x) which tells limit doesnt exist . But answer is 0 . i have verified with graphical calculator ... What am i doing weong here ??? i have no clues ...

It could be correct, but could it be made without LH?

What about x[(1+1/x)^x - (1-1/x)^-x] with x--> infinity? Is It possibile to solve without derivates?

Soooo much better than football!

Hey dude, is there any way I could support you and your content by something like a Patreon account ?

Me: people will think you're up to something

This is very counter intuitive!!

Why can you differentiate by replacing n with x when it's natural?

4:55 I see a stick man and 1/x

11:54 "I am suspecting, 為什麼這個是..."