Math for fun#5, MOST SKIPPED LIMIT PROBLEM ON CALC EXAMS

Math for fun#5 (calc1), HOW CRAZY IS YOUR LIMIT!
More math for fun: • Playlist
Limit as Riemann Sum,
limit as an area,
definition of integral as a limit,
integral of 1/sqrt(1+x) from 0 to 1,
blackpenredpen
follow me: / blackpenredpen

[@ijabbott63]

8:10 I'm wondering why the upper bound of the summation changed from n to ∞. Was that a "typo" (or a "writo")? I guess it doesn't matter as you're trying to find the limit as n→∞ anyway. Cool video, by the way! I love your teaching style.

[@blackpenredpen]

Ian Abbott I just noticed that too. Yes you are right. It should still technically be "n" unless I don't have the lim in front any more

[@barutjeh]

Good spot; it does actually matter, as for any fixed n, that infinite sum is divergent, meaning the limit over n loses it's intended meaning. In other words; it's important the bound for i and the value of n go to infinity together, otherwise wild stuff happens. If n goes to infinity first, the whole thing ends up being 0. If the bound goes to infinity first, everything diverges.

[@reazraza]

Ian Abbott yes he wrote it beacuse n tends to infinity. But he forgot to mention it

[@myself0510]

@@reazraza No, it really has to be n and then set n to go to infinity, since n appears in the summation as well. Otherwise each term of the sequence is an infinite series that goes to infinity

[@akasin8202]

@@reazraza in case the question asked for I from 0 to 2n the integral would have been from 0 to 2 ....so it does matter what he writes for summation limits..he has to write "n" specifically

[@chrismanich3063]

i can't believe i am watching this at 2 am and seeing you solve this making me happy, nice job. I hope one day i will be as good as you at maths :D

[@blackpenredpen]

Chris Manich thank you! Seeing your comment makes me happy too!

[@Hexagy]

You make the "Lim->N" "Lim->X" in your thumbnail

[@user-en5vj6vr2u]

*l i m i t a s l i m a p p r o a c h e s N*

[@paradoxica424]

A more unconventional method: By judicious squaring, it can be shown that: √(n+k+⅔) - √(n+k-⅓) < 1/(2√(n+k)) < √(n+k+½) - √(n+k-½) The bounds clearly telescope upon summation. Denote the middle sum to be S The limit we want is 2S/√n Summing both sides, we obtain: √(2n+⅔) - √(n+⅔) < S < √(2n+½) - √(n+½) Then it is clear that S/√n tends to √2 - 1 by the squeeze theorem. Thus, the answer is 2√2 - 2

[@andrewstallard6927]

Excellent That method is more rigorous than the one in the video.

[@Mewtwo332]

What makes you think the method shown in the video isn't rigorous ?

[@riccardoorlando2262]

He passes through some undefined limits, and non-converging series.

[@riveraeduardo3184]

Riccardo Orlando What the? brotha, the integral he generated, which he proved to equate to the original limit problem, runs from 0 to 1... it all checks out.

[@thatdude_93]

could you please explain this further? what do you mean by judicious squaring?

[@alkankondo89]

This is an excellent example of how it is easy (at least for me) to misjudge the simplicity of these limits. My immediate reasoning was "well, each term goes to 0 as n goes to infinity, so the whole limit must be 0." But when he explained that adding up an infinite number of 0's in limits is not necessarily 0, I realized that this is an example of the indeterminate form, 0*infinity. Who would have guessed that this limit would be something so arbitrary as 2sqrt(2) - 2? Very strange!

[@NoNameAtAll2]

0+0+0+....+0=0*inf, which needs to be looked into

[@iabervon]

My quick check was that each term is between 1/sqrt(n)sqrt(n) and 1/sqrt(n)sqrt(2n). So for any n, the sum must be between 1 and 1/sqrt(2). So the limit isn't either 0 or infinity.

[@allaincumming6313]

This case is pretty much as 1^inf, where if "1" isn't given by a based constant 1 in the function on the limit , it's not necessarily 1.

[@christianrodriguez823]

Wow i can't believe I got this right...I guess there IS a benefit to having a hard-ass professor for a whole semester :P

[@blackpenredpen]

Christian Rodriguez love your comment! you must have a great understanding! Keep up the good work! If u can, check out my fb at facebook.com/blackredpen and u will see ur comment as my profile pic. Lol

[@christianrodriguez823]

blackpenredpen Thank you so much! I haven't felt this honored since I gave a speech at my high school graduation lol

[@geico105]

I’m not going to watch this video because I want to eventually figure this out for myself.

[@leif1075]

@@blackpenredpen yea but intuitively the limit of a sum equals the sum of the limits and each of these goes to zero...

[@nath8123]

Sooo awesome! Absolutely loved it!

[@blackpenredpen]

Natália Rodrigues cool! Thank you!!

[@riveraeduardo3184]

This is one of the most exciting math videos I've seen! That whole fast motion during the u sub integration had me laughing out loud! I love that I found this channel and I love the energy you have in your lectures. This channel is bound to blow up in subscriptions fast! Thank you!

[@Cannongabang]

loved it even though i still see integrals as "far more abstract" than areas of rectangles. I see derivatives and understand them, integrals are like strange

[@blackpenredpen]

Cannongabang lol. Integrals are cool too

[@TheNormMan]

on thumbnail it's "x->∽"

[@wardcampbell5885]

blackpenredpenbluepen!

[@_patrickbrophy]

I see the Supreme. Hypebeast and a math wiz, who knew.

[@michalchik]

why is he holding a thermal detonator in his left hand? 😉

[@glenn1you0]

michalchik i thought he was an Ood from Dr Who

[@riveraeduardo3184]

michalchik because he's Blackpenredpen

[@rainbanreborn1533]

why r u copying comments

[@KnakuanaRka]

A, that’s a microphone. B, I could swear I’ve seen that exact comment elsewhere.

[@user-mt9ux2di6u]

It's a mic

[@eruditeroach7099]

Wow, I was expecting this to be a really upper-level calc problem, but it can be solved with just Calc 1 skills!

[@emanuellandeholm5657]

The terms are not terribly nonlinear as n tends to inf. An approximation of the partial sum would be 1/2 (1/sqrt(n^2+1)+1/(sqrt(2n^2)) ~ n/2 (sqrt(2) + 1) / sqrt(2n^4 + 2n^2) ~ n/2 (sqrt(2) + 1) / n^2 sqrt(2). Sum n of these to get n^2/2 (sqrt(2) + 1) / n^2 sqrt(2) = (sqrt(2) + 1) / (2 sqrt(2)) ~ 0.85355 The correct answer is ~ 0.8284, so as expected we're not that far off. I really liked your transformation of the partial sums into a Riemann sum, very instructive!

[@PeppoLS]

So elegant solution!

[@vameza1]

Wonderful. This limit opened my eyes!!!

[@srikanthtupurani6316]

i think we can calculate the limit without using integrals and fundamnetal theorem of calculus by studying the growth of functions (1+r/n)^(-1/2). we need to use binomial theorem.

[@daviddevoogdt4497]

another solution: call serie up to n: P(n) if we assume there is a solution to this problem then: * P(n) = P(n+1) at infinity * sqrt(n)*P(n) - sqrt(n+1)*P(n+1) = 1/sqrt(n+1) - 1/sqrt(2n+1) - 1/sqrt(2n+2) (simple algebra) after working out limit one finds P(n) = 2*sqrt(2) - 2

[@yoyoezzijr]

Can also be the integral of 1/√x from 1 to 2 if we set 1 + i/n to be the a + i(ΔΧ) and its pretty cool because the u-sub does exactly that

[@jorgefiestas9542]

Intresting question, beautifully explained!! keep up the awsome work!

[@fengshengqin6993]

Thanks for BlackPenRedPen ! I have figure it out in 5 mins before I check with his right answer.

[@ikarienator]

Let C_n = 1/Sqrt[n] Sum[1/Sqrt[n+i],{i,1,n}]. If the sequence converges, we get lim C_n = lim C_(n+1) which gives lim C_n = lim (1/Sqrt[2n+1] + 1/Sqrt[2n+2]-1/Sqrt[n+1]) / (Sqrt[n+1] - Sqrt[n]). Sqrt[n+1] - Sqrt[n] is asymptotic to 1/2 * 1/Sqrt[n], so lim C_n = 2(Sqrt[2] - 1).

[@AntoshaPushkin]

Wow, it took me 5 minutes to figure this out, calculus at the place I'm studying kicks ass! Well, I took it the previous semester but I still feel nice for solving this thing so fast, and I did it exactly the way it was presented

[@user-vp7ly3ih3n]

Где ты учишься?

[@AntoshaPushkin]

@@user-vp7ly3ih3n в Израиле, Технион. Когда я написал этот комментарий, это был следующий семестр после курса "мат. анализ 2"

[@user-vp7ly3ih3n]

Антоша Пушкин Сейчас я так понимаю ты окончил учебу?

[@AntoshaPushkin]

@@user-vp7ly3ih3n угу

[@devroopsaha4020]

MUCH EASIER THAN SQUEEZE THEOREM, THNX BPRP

[@joseluizdurigon8893]

That was beautiful. Magnificent. I loved it!!!!

[@thayanithirk1784]

Please solve some very tough classic problems from ittjee

[@Jesin00]

I usually use k for the index when n is taken, so nobody confuses it for the square root of -1.

[@Smiling_Tears]

Such an elegant solution!

[@Actanonverba01]

Love this one! Great job explaining it.

[@acorn1014]

Haven't taken calculus in 3 years. All I could do was prove that the limit was somewhere between 1/2 and 1.

[@alexli6935]

It shows that you are willing to tackle problems and that you actually understood what it was asking, very good

[@gongasvf]

Don't really get why the upper bound of the integral is 1, it seems to me that we could've said it is any number. The fact that one is the numerator of the fraction in the limit doesn't seem a good justification for the upper bound of the limit to be one... (imo)

[@dlevi67]

The integration interval is [0 1] by construction, precisely because it simplifies your life later on.

[@mariomario-ih6mn]

You brought back your intro at the end of the video.

[@damirbajramovic292]

You guys rock. Really helpful. Keep up the good work!

[@blackpenredpen]

thanks!!!!

[@6612770]

Finally, I've seen it done! Have HEARD that a summation problem can be turned into an integration problem, but that's all. My uni course must have been deficient. :-(

[@Linvael]

The good old "solving the problem by knowing the solution and finding a way to get there".

[@blackpenredpen]

Linvael and by experience

[@NoNameAtAll2]

most of math comes from finding some solutions beforehand and than finding where they're applied

[@mostafaahmednasr621]

Best tging about math problems is the ending

[@BenjaminKeilty]

I saw it at 8:10 and I stared in amazement for a solid minute. Wow.

[@krisneyens1272]

I genuinly liked this video in particular. Grand!

[@blackpenredpen]

Kris Neyens thank you.

[@willie333b]

I remember seeing it in high school

[@zorm_]

Hello ! never learned to calculate integrals that way... still looks super cool ! I would have calculate the anti-derivative of 1/√(1+x) which is 2√(1+x) and yeah I like it !

[@infirmuxx]

Brilliant.

[@mohammedalashker1105]

Very nice thank you

[@Rehbock137]

This seems like an artifact of the method to take a limit due an infinitesimal being greater than zero.

[@chritophergaafele8922]

the index of summation goes from 1 to n

[@mmatt314]

love the high pitched "right?"s!

[@blackpenredpen]

Matthew Shepherd right?!

[@Treegrower]

I have a few questions. First, how did you change i/n into x? What makes that ok to do? Also when you were integrating at the end you changed the bounds of the integral from 0 to 1 ---> 1 to 2. Why and how?! Thanks, cool video!

[@blackpenredpen]

1st. Notice if f(x)=1/sqrt(1+x), then f(i/n)=1/sqrt(1+(i/n)). And in the vid, I just looked/worked this backwards. 2nd. That's an excellent question, why NOT from 1 to 2, right? And the answer is, IT COULD BE. but then your function would be 1/sqrt(x). And the reason is the area under 1/sqrt(1+x) from 0 to 1 is the same as the area under 1/sqrt(x) from 1 to 2 And likewise, you could also use a different function and set it as from 3 to 4, etc. I used the easiest one, which is the area under 1/sqrt(1+x) from 0 to 1 Hope these help.

[@Treegrower]

Ah, very well explained. Thank you!

[@GermanSnipe14]

Hey, I know I'm 5 months late but I think you were referring to the integration he did in blue/red (when he sped up). I'm just clarifying because I think he understood your question a tiny bit differently. He changed the bounds because of his u-substitution -- when he said "u=1+x" he is integrating with respect to u, not x anymore, so he must change the bounds accordingly. The lower bound is x=0. Plugging in 0 for the u equation, u=1+0=1. So our new lower bound is u=1. Plugging in the upper bound, x=1, we get u=1+1=2. So our upper bound is u=2. I think his answer indirectly answered your question but if you were just asking why he was able to do that, this is why. :D

[@reazraza]

ᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟ ᅟᅟᅟᅟᅟᅟᅟᅟ ᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟ ᅟᅟᅟᅟᅟᅟᅟᅟ ᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟ ᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟ ᅟᅟᅟᅟᅟᅟᅟᅟ ᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟᅟ thats basic integral principles watch his other videos or read a book. You will find out

[@marshmellominiapple]

yea I'm not in calc so you speeding up that last part went over my head

[@vidyutawasthi7814]

I though this video is of BlackpenRedpen, but it turned out to be of BluepenRedpen!!!

[@2010RSHACKS]

Very cool! I never had anything like this in calculus class and I would've gotten it wrong if I did.

[@littlejimmyxD]

Awesome video

[@blackpenredpen]

Cheesyfangs thank you

[@snakespeak]

Totally awesome.

[@mikedavid5071]

Love the background music.

[@DanBurgaud]

i got lost somewhere in the 10th minute mark..... I need a stronger coffee...

[@trumanknight873]

Great channel!

[@albendsouza]

Limit of a sum where tn is 1/root (1+ (r/n)) that will just be integral 0 to 1 1/root(1+x) dx so 2(root(1+x)) and substitute limits Edit: i type the answer after clicking on the video

[@papapapapapapageno]

I like it! Cool.

[@bensrotation7021]

Thanks for the video.🤓

[@Seki-Saed21]

my teacher thanks

[@blackpenredpen]

You're welcome!

[@dalisabe62]

Not convinced of the last minute hasty comparison. Both i an n go to infinity, so the quotient is undetermined. What is n to start with? Is it a natural number, integer or real? Assuming it is a natural number, we could start with n=1,2,3,...1000 and see what the sum evaluates to. It is obvious that the sum is monotonically increasing, however at a decreasing rate as the denominator is getting bigger on every term of the summation. If the limit found in the video is correct, we should get a sum that is always less than such limit. We need a software like Mathematica to do the work. In particular, comparing the infinite sum with the integral over [0,1] interval needs more justification.

[@nellvincervantes3223]

Love your background music.

[@sandmann6851]

Perfect one

[@skya6863]

I just looked at it quickly and assumed it was sqrt(2)÷sqrt(3) but it was 0.01 off

[@skya6863]

Took a more serious approach to the problem and managed to solve it. I made an integral that had upper bound n and lower bound 0 with the function being 1/(sqrt(n)*sqrt(n+x)) dx Integrated it and got [2sqrt(n+x)/sqrt(n)] 0,n No matter which n you choose this becomes 2sqrt(2) and the 0 becomes 2. So 2sqrt(2)-2 which was the answer

[@hubenbu]

This is truly sweet and juicy, I love it!

[@tomasbeltran04050]

This looks like a telescope

[@Abdega]

11:08 Child Stars Where are they Now?: Simon the chipmunk

[@typo691]

Nice

[@Dunkle0steus]

haven't done calc since first year and its all coming back to me... had to slow down the vid to 0.5x on the sped up part to make it. Sounds a little chipmunky and still fast, but I could make it out.

[@MrPhilipp456]

Great vid man! keep it up

[@blackpenredpen]

Philipp Huber thanks!

[@natealbatros3848]

Why the limit of the integration is 1? It couldn't it be any positive number?

[@martinepstein9826]

How could it be greater than 1? The sum doesn't include any f(x) with x > 1

[@douglasstrother6584]

Slick!

[@spudhead169]

Wow, that was hardcore.

[@elbisho2045]

😭😭😭 i wish you could be my teacher

[@blackpenredpen]

Vicente Valdés I am! On KZbin!

[@blackpenredpen]

Vicente Valdés thank you for watching

[@GammaDigamma]

You sound best in fast forward...

[@trucid2]

You have this question on a calc test? That's playing dirty!

[@jatarokemuri4549]

on the thumbnail, I thought it was obvious. the answer is the sum itself as the limit uses x, but the expression uses n.

[@PhilippeFrancais85]

Beast Lad!

[@blackpenredpen]

PhilippeFrancais85 ?

[@PhilippeFrancais85]

blackpenredpen You're a beast at math. It just means you are very good. Great videos - keep it up!

[@darioniero2641]

i tryed this limit yesterday evening and i didn't get an idea of how to solve it. Your exeplanation is very brilliant. Love this limit. Keep it up. Love from italy!

[@poesie_filosofiche_poems]

The x is build as i/n so It is only the rational numbers between zero and one the irrational numbers are not involved. So the exercise is wrong.

[@wilmermoralesmatematicas3440]

Very good my friend crack.

[@abhishekchoudhary4689]

Hey this should be possible with sandwich 🥪 threoum ( squeeze) and using sandwich we get the ans 1. Which is not matching the real ans why????????

[@DarthChrisB]

I paused the video and wrote some notes with my mouse in MS Paint. Wow, my Paint sheet is fairly empty compared to your white board and I got the same result. You didn't need that 1/√(n)√(n+i) stuff, all you need is 1/√(x), then integrate.

[@blackpenredpen]

DarthChrisB right. I was teaching, you were just solving.

[@ranjiniravishankar6832]

Hello! This is a great video, and a cool problem. But I am still not convinced why adding infinitely many zeroes will not give a zero? Could you expand on that? Google is iffy on this subject atleast. Thanks!

[@HipposCanFlyy]

if you think about what we mean by '0' in limits, it is actually really an infinitesimally small number.

[@ranjiniravishankar6832]

I've understood by now, I forgot to update my comment here :)

[@tilmanroeder609]

Ranjini Ravishankar When you take any number n element N, you will get a perfectly reasonable result s. If you now take any number r element R, you can find a result such that |2*sqrt(2) - s(n)| < r, by choosing a sufficiently large n. Hence as n approaches infinity, the limit approaches 2*sqrt(2)-2

[@johnniejones8007]

The thumbnail has the limit as ‘x’ approaches infinity instead of ‘n.’

[@tollboothjason]

11:07 you just became a chipmunk!

[@ajaypahadiya824]

Believe me . I did it without pen paper

[@longsteinpufferbatch4949]

I did it without pen and paper in half a minute lol. It's nothing huge

[@reazraza]

This guy is very good he will make a great professor

[@blackpenredpen]

Thank you, but I am a teacher already : )

[@reazraza]

Oh sorry. Hey can you make a video of finding a general equation to predict a fibonacci number. It will be a great help

[@ianprado1488]

Hey you make great videos.

[@manamimnm]

thank you sir

[@blackpenredpen]

manamimnm you're welcome. Thanks for watching too

[@pedromalafayabaptista3655]

Hi. First of all i want to congratulate you for your videos. Top-notch work. Secondly i would like to propose an easy, but i think funny, problem. Let´s call it blackcardredcard: I have a 52 cards deck that i break into two subsets; i know that on the left substet there are 3 times more black cards than red cards; which is the SMALLEST number of black cards i must take out of the right subset to have 3 times more red cards than black cards there?

[@carljohnson1082]

Very helpful vid mate thanks. Subbed to you in hopes of more great videos. Some constructive criticism if you will, perhaps lower the sound of your outro or change it, it was pretty obnoxious xD. The music in your video was a peaceful melody but your ending :S Great video either way

[@blackpenredpen]

Carl Johnson hi Carl, thanks for the comment. I watched the end again carefully and would like to agree that I need to lower the volume. The kids voice was too loud in comparison with how I ended my lesson. Lol. Thanks again and will have more MFF vids later on

[@sivad1025]

That is beautiful

[@disciplinenepal5081]

good from nepal

[@pranayvenkatesh8815]

Good problem, but I found it quite easy. There are some harder summation questions where finding the general form of every nth term is difficult. Converting sigma to integral is easy though. IITJEE problems are the bosses at killing us with this.

[@General12th]

Do you think you would have gotten this answer right when you were taking calculus classes for the first time? Or would you have skipped it, too?

[@mokouf3]

I wonder if this is really most skipped.

[@miguelangelordonez6957]

Beautiful :3

[@destroctiveblade843]

a problem I found when I tried to find the limit on my own before watching the answer is that I started from 1

[@WindsorMason]

1/sqrt(2) ≈ 0.7070, and 1 is obviously = 1. And now the exact value is: 2*sqrt(2) -2 ≈ 0.8284. So the exact value is within your bounds.

[@destroctiveblade843]

oh yeah hhhhhhh that was a mistake XD

[@WindsorMason]

Hehe, good luck and take care!

[@op13579nv]

Great video. The presentation is very neat. Thank you!!

[@jatinderkaur7993]

Why can't you apply sandwich theorem

[@user-ph2kw6jf7b]

Why does x go to infinity, not n?