A COUNTER-INTUITIVE CALCULUS LIMIT

[@josephl6896]

Casually doing math while holding a thermal detonator.

[@warlockgnome1015]

I'm so glad I wasn't the only person thinking this

[@prateekbhurkay9376]

If he stops doing math, the thermal detonator goes off. It's like the movie Speed.

[@lswcs]

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 :)

[@samueljehanno]

Nice

[@AlRoderick]

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?

[@blackpenredpen]

Alexander Roderick lol!!! Nice one

[@zabrakhan5178]

+blackpenredpen Hello! Can you make a video on how to solve the integral of this (5x^(2)-x+47)/(x^(3)-x^(2)+16x-16)? But can you do partial fractions with the imaginary roots?

[@JayJay-ej6dq]

blackpenredpenbluepen

[@moregirl4585]

but there's no hyperlink on the video xd

[@munendersingh5631]

Below

[@altacc7836]

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

[@timtimtimmaah]

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

[@fxyang1989]

Good thinking. It starts from 2 when N=1 and e>2.

[@justarandomcatwithmoustache]

Isn't it obvious ?

[@gljames24]

It makes sense as the part -ne is starting at e, while the part n(1+1÷n)ⁿ has to reach e as it approaches ∞, so it is starting later.

[@masongatz2810]

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

[@FernandoRodriguez-et7qj]

Who are you mason? You look like a professor

[@General12th]

@@FernandoRodriguez-et7qj You are YOU, Fernando? You look like a professor too.

[@krukowstudios3686]

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

[@yuklungleung620]

Krukow Studios i would say you are on9

[@2ndtik]

@@yuklungleung620 saupei la sohai

[@modemanslutning]

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.

[@jordank195]

Ah I see the KZbin gods also presented you with a 5yr old video 😂

[@leandroteles7857]

You can also think of it as a product of 2 functions: One part is the product of x by the constant function "e". The other part is the product of x by the function (1+1/x)^x. This function is NOT constant, and in fact, it only reaches "e" at infinity. Therefore, it is always smaller than the constant function "e".

[@FlyNavy906]

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

[@paul_w]

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.

[@ramez2775]

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

[@PackSciences]

That moment when you forget a minus sign

[@blackpenredpen]

PackSciences so true...

[@tollboothjason]

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

[@FernandoRodriguez-et7qj]

toll_booth that wasn’t funny

[@FernandoRodriguez-et7qj]

You made me rewind the video

[@hugolaurent2643]

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 ;)

[@Aditya_196]

🙂 can u tell me what's Taylor expansion and how did u use it here ?

[@gillrowley]

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

[@dugong369]

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.]

[@FuhrerShattercore]

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

[@General12th]

I love this channel! It's like the BPRP of KZbin!

[@xavierplatiau4635]

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.

[@deinauge7894]

same :) at least very similar to what i thought while preparing a meal... ln(1+1/n) ~1 + n^-1 - 1/2n^-2 ln(first term) - ln(second term) ~(2n)^-1 and with the derivative of ln at en: ln'(en)=(en)^-1 the limit is ((2n)^-1)/((en)^-1) = e/2 higher orders in the expansion of ln vanish obviously ^^

[@Aditya_196]

😅 can u explain it ?

[@restcure]

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!

[@blackpenredpen]

restcure maybe!

[@veradrix]

But he can only apply L'H to function fractions, and functions are in terms of x not n, so he did it right!

[@ianmoseley9910]

Would still have worked if he used d/dn - nothing sacrosanct about using x.

[@ianmoseley9910]

Irina Alexandra functions can use any letter as a variable - if you run out of. letters there are other alphabets

[@racheline_nya]

@@veradrix it works with any symbol instead of x. you just need to replace every instance of x with that symbol, including the one in d/dx. so d/dn(f(n))=d/dt(f(t))=d/dx(f(x)).

[@m.guypirate6900]

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.

[@ryanchatterjee]

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

[@vvalph9483]

70% of the time i was looking at his beard

[@calyodelphi124]

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

[@fredsharp7419]

Brilliant! Your enthusiasm is infectious!

[@matchedimpedance]

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.

[@matthewzuelke6721]

4:30am. Good morning, time for BlackpenRedpen

[@blackpenredpen]

Matthew Zuelke WOW!!!!!

[@tiboletibetain5730]

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

[@haydenkarkainen1167]

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.

[@JustSimplySilly]

You make limits interesting!

[@Anas-nu7io]

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!

[@physicsphysics1956]

The integral is likely non-elementary.

[@reeeeeplease1178]

Kinda late to the party, try u=1-cos(x) and the integral *should* turn into 1/sqrt(2) * sqrt(1-u) du The rest is quite simple

[@evgeniyevgeniy8352]

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.

[@blackpenredpen]

Evgeniy Baton Ahhh, yes your were right. What I said at 4:20 was bad. But I think the work is still legit. The theorem is: if you have a function such that f(n)=a_n, then we will have if limit of f(x) goes to L as x goes to inf, then a_n goes to L as well. But the converse isn't true, just like your example. So, when I checked the limit as x goes to inf and I got the answer, then I was okay for me to conclude that a_n goes to the same limit. I will make a video on this. Thanks.

[@blackpenredpen]

Video coming soon! : )

[@nath3x]

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

[@mr.soundguy968]

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

[@jinanlife]

the lady at the beginning is absolutely gorgeous

[@lucasscoz6090]

math?

[@Myuri3146]

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

[@danielbenyair300]

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

[@douglasmagowan4918]

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.

[@mihaiciorobitca5287]

You and Peyam are the best

[@SJ-ry6br]

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

[@thinhtran3593]

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

[@someperson188]

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)

[@BabyXGlitz]

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

[@mrflibble5717]

Keep up the good work, enjoy your Channel!

[@qbert8695]

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

[@indrada-rf2vu]

In Dev I got the same result using a calculator

[@Mayank-mf7xr]

the lovely doraemon sequence . love you man

[@royler8848]

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

[@forcelifeforce]

* you're -- It's for "you are."

[@mtaur4113]

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

[@xiang-yue-fung]

"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.

[@georgepapamichael590]

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

[@levkrainov]

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.

[@duranchen3222]

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.

[@forcelifeforce]

That would be 1/x - 1/(1 + x).

[@Czeckie]

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

[@anshumanyadav24816]

Best video ever made by you 😛💕

[@profesordanielalvarez3498]

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

[@barryzeeberg3672]

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?

[@AayushSrivastava0307]

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

[@frenchcoconut]

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 !

[@maxencen.9550]

et honnêtement les comparaisons de suite sont bien plus élégantes

[@jeromemalenfant6622]

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.

[@ozzyfromspace]

Actually that generates an e*(inf - inf) type situation again. But for what it's worth, I considered that too for a few seconds before realizing this. Best wishes friend 🙏🏽.

[@112BALAGE112]

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

[@ponirvea]

He used it four times; at 11:04 the limit evaluated to -1 by applying l'Hôpital's rule twice more

[@monzur1947]

Ahmad Fares that's not using l'Hôpital's. It's just using the fact that all the terms of x with power less than the highest become insignificant as you go to infinity.

[@JensenPlaysMC]

What rule is this btw? for the insignificant values of a power less

[@carly09et]

This is the same as turning it into a sequence and using Supersums watch a Numbephiles' video before this the logic is the same. chain l'hopitals ==> Supersum :)

[@agfd5659]

@@JensenPlaysMC if you want it more formally, I would factor out the x^2 from the top and from the bottom, reduce them and notice that the top is equal to -1 and the bottom approaches 1, so you can take the limit without a problem now.

[@reetasingh1679]

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?

[@Mar184]

Yup, there is no way to accurately evaluate (1+1/n)^n for large n with ordinary floats. You could import the "fractions" module to utilize Python's arbitrary precision integer arithmetic by letting it treat it as a fraction - then you have perfect precision for that part, but of course the runtime will explode completely and quite quickly instead. And you still don't get any further since you can't express "e" as a fraction, so the e*n part will become inaccurate either way. A more promising approach is the "decimals" module. You just have to set a custom, really high amount of decimal digits. for "d" decimal digits, you should be fine for something up to just shy of n = 10^(d-1).

[@reetasingh1679]

Mar184 Thanks for the detailed answer! I will surely be checking out the 'decimals' module

[@reetasingh1679]

Doggy Jr Actually no, the answer is definitely -e/2. Like I said a month ago, I tried writing a program to test the limit because I was unconvinced too. But remarkably the program gave results close to -1.35 for values of n as large as a billion.

[@italolima855]

Put It in a Scientific Calculator with n=10^9

[@reetasingh1679]

Doggy Jr Fully agree with you... But that wasn't my issue. My issue was with accuracy. The limit worked perfectly for relatively high values of n. At a certain point the limits started to blow up. I get that the number we are storing in the program is finite and not exactly equal to e. But we could still use a whole lot of decimal digits of precision to achieve the answer.

[@xauriclegaming8654]

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

[@cameronspalding9792]

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

[@alre9766]

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

[@sohammakim9178]

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

[@zahari20]

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.

[@GunSinMovies]

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

[@jceepf]

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....

[@GDPlainA]

Me: people will think you're up to something

[@tonyhaddad1394]

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 !!!!!!!!!!!!!!!

[@johnnicholson8811]

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

[@zohar99100]

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.

[@tonypalmeri722]

I liked this, because ultimately you convinced me of something that was very counter-intuitive at first. I gained an important insight by watching this and thinking it through. The final convincing came, though, when I went to an online graphing tool and graphed that function, out in the area of x around 5000. Yes, I *know* that graphing is not the way to get the actual limit, but if something radically different was going on around x=5000 I would have been suspicious of the result and investigated further.

[@chasemarangu]

0:03 at n=∞ (1+1/n)ⁿ=e but did you know that at n=φ (1+1/n)ⁿ=φᶲ?

[@chasemarangu]

1:35 I had absolutely no idea so I did it on a graphing calculator and it appears to be -e/2 but let's think about this. Also I noticed on the graphing calculator that x·(1+1/x)ˣ−e·x is same as x·((1+1/x)ˣ-e) by the way lim (x→∞) all the time and so substituting already we get ∞·0 which means it could be anything on the interval [-∞; ∞] so we need to do more work, I am gonna do something really stupid now. Using the negative exponents we get x/((1+1/x)⁻ˣ) − e·x and we can do the really stupid thing now that is use L'Hopitals rule because I am calculus novice and don't really know what else to do so then we get 1/something − e·x and then you find out on wolframalpha that that something is too long to write out so you give up

[@chasemarangu]

Actually, after watching the next ~45 seconds of the video it turns out I wasn't too far off you do use L'Hopitals rule lol only I did the wrong multiplicand switching to the denominator also I messed up because I didn't keep x times everything instead I did x·something − x·e well oh well

[@chasemarangu]

in the end I am glad I did not do all the math that is a long time for one problem and my graphing calculator (aka cheating) guess was right

[@returnexitsuccess]

Really cool video! Very surprised to see Santa Barbara show up though, come visit UCSB :)

[@blackpenredpen]

returnexitsuccess I was actually there two weeks ago just for sightseeing. It's very pretty

[@Djake3tooth]

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…

[@jayantimajumder4151]

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..

[@aymaniklayli298]

Nice solution and prefect you sir

[@ЮрійЯрош-г8ь]

Thanks for the video. Why can't we using product and sum rule for limits show that it is equal to zero ?

[@MuffinsAPlenty]

The product and sum rules for limits only apply when all of the limits you're splitting things up to exist and are finite in value. The problem with this situation is that the limits you want to split it up into are not finite. Both "terms" have a limit of infinity. You could recognize it as the indeterminate form ∞−∞ or you could factor out the n like blackpenredpen did and recognize it as the indeterminate form 0∙∞.

[@ЮрійЯрош-г8ь]

MuffinsAPlenty oh, yeah, I understood, thanks for the answer.

[@megacarlosloki]

Hey can you make some videos solving ITA-SP questions?

[@proclubsjov3583]

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

[@Bicho04830]

11:37 You desappeared at the last frame. O_0

[@danielrosado3213]

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!?

[@robertj4424]

If you use substitution x = 1/n, then you can get the final answer quicker: lim{x->0} [(1+x)^(1/x) - e]/x = lim{x->0} (1+x)^(1/x) [(-1/x^2)ln(1+x) + 1/x - 1/(1+x)] = lim{x->0} (1+x)^(1/x) [(-1/x^2)(x - x^2/2 + 1/x - 1/(1+x)] = lim{x->0} (1+x)^(1/x) (-1/2) = -e/2

[@sinom]

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

[@supercool1312]

Sinom its not, because thats just 0

[@kaizoisevil]

@@supercool1312 If this is a limit where n goes to infinity, then 0*n cannot be simplified to 0.

[@infirmuxx]

Good math, as always.

[@wolfram9669]

Very good video! Thank you.

[@tomasthemas]

In your final Limit, why do we omit the 2x+1 on the bottom? Doesn't that meant the bottom approaches infinity faster than the top?

[@danielmiddleton6094]

Technivally yes, but think about this. Infinity^2 is a lot bigger than 2 (infinity). Infinity times bigger, in fact. So it effects the outcome by basically nothing, so we can "essentially" get rid of it, leaving the x^2

[@OonHan]

OMG So interesting

[@pasawatviboonsunti9066]

Plus e/2 in the limit, and multiply by n, then you get another limit to find.

[@PsyKosh]

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

[@blackpenredpen]

n is usually for whole numbers, so d/dn wouldn't make sense.

[@87我是-h3u]

Psy Kosh *usually*

[@PsyKosh]

Go wild and differentiate with respect to n anyways. I won't tell. ;)

[@pierreabbat6157]

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.

[@jarikosonen4079]

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

[@shezanahmmed5582]

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

[@NO-ll4fp]

I never thought when n tends toward infinity that basically e times n minus actually e times n would be negative e on 2

[@ahmedalmerza1707]

So then is the limit form of "e" is actually an underestimate of "e" itself, given that the solution is a negative number?

[@kaizoisevil]

Yes, it approaches e from the left.

[@hydraslair4723]

@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?

[@cecilhenry9908]

This is very counter intuitive!!

[@blackpenredpen]

Cecil Henry it is!!! I first thought it was 0

[@marcelweber7813]

And now everything makes sense and last weeks video weren't just for fun.

[@ЭльЯвор]

no problem it's equel to one obviusly