Don't use L’Hopital

[@AndDiracisHisProphet]

wow, the last one....that was subtle. thank you for that

[@johannesmoerland5438]

The last one depends on your definition of sin and cos, in Analysis, we defined the functions as Im(exp(ix)) and Re(exp(ix)) and the proof was valid

[@davidgould9431]

I was wondering that, too.

[@paolokaibrennan6514]

So was I

[@beardedboulderer2609]

If you work out the proof of this, you need the taylor series expansion of sin and cos. To derive those, you need the derivative of sin and cos.

[@beatoriche7301]

@@beardedboulderer2609 Not if you _define_ sine and cosine to be the real and imaginary parts of the complex exponential. Most analysis courses don’t define sine and cosine via the unit circle as high school math courses do, but either via their Taylor series expansion or the complex exponential. And since it’s pretty easy to show that the power series defining sine and cosine have an infinite radius of convergence, they can be differentiated term by term (that’s a famous theorem in complex analysis; complex power series *always* converge within a circular subset of the complex numbers - though this circle may have infinite or zero radius -, and for all complex numbers within this circle, the function given by the power series is differentiable with its derivative being equal to the series obtained by differentiating term by term), which immediately yields their derivatives without requiring the limit in question at all. A lot of the time, there is no one fundamental way to define things in math; Euclid’s lemma in number theory takes considerable effort to prove going by the usual definition of primes, but sometimes in abstract algebra, it is more convenient to define prime elements directly via Euclid’s lemma, and then it’s just the definition of a prime. And when you get to higher levels of mathematics, you enter fields where, more often than not, people have spent decades fleshing out the definitions easiest to work with. If you go by the naive definition of sine and cosine via the unit circle, the limit in question is certainly not trivial. However, while there is a good reason high schoolers aren’t immediately confronted with the power series definitions of sine and cosine (despite it being much more powerful, generalizing seamlessly to complex numbers and even matrices), I haven’t seen a single analysis textbook that defines trig functions via the unit circle; it just makes things much more complicated than they need to be, and you’d have to flesh out a full theoretical apparatus to even make the definition of these functions rigorous, since, if you start doing analysis, it’s not at all clear what an angle or the length of a curved line mean (that already requires quite a bit of measure theory).

[@samueldeandrade8535]

I would say you are mot being very clear. Because your sine is NOT the sine of the problem. At least, we don't really know that immediately, do we? By your definition, I understand that lim_{x→0} Yoursin(x)/x = 1 is pretty much automatic. But Yoursin(x) just uses the same name as that sin(x). You would have to show they are the same.

[@renzalightning6008]

What you said at the end there was echoed by one of my lecturers when I was doing Epsilon proofs using limits, he said you're assuming what you're trying to prove instead of working through the algebra and ending up with the result. It is a very powerful tool.

[@musik350]

3:45 that would indeed really be insane, even according to Einstein's definition of insanity

[@juliansanchezcastro.2442]

I'm taking a course of Lineal Algebra, and with the second the first I though was "well, in the infinity the x square dominates the 1, and the it's cancelled with x bellow". You are right, we get used to think in only one way with this kind of things. Thanks! :D

[@nanigopalsaha2408]

I substituted x = tan θ

[@thomasborgsmidt9801]

That is definately one of Your better video's! It makes its points succinctly and without unnecessary embelishments. It attacks problems of understanding clearly!

[@szegeretcl9318]

Thanks for staying happy in every video. Your content is great.

[@pedrocrb]

Great! I would love to see more examples where using some well known theorems get you in trouble. Like "dont use integration by parts!" or a "dont use x" series

[@advaithbala]

Me: worried to discover that a useful piece of math might be wrong Also me: Realizes math is perfect and consistent and the video is actually about how not to use said piece of math incorrectly

[@bullinmd]

Blackpenredpen has covered the circular reasoning with sin x / x.

[@fredrikrenstroem1661]

Loved it, learned about l'hospital fr today. Subscribed.

[@japotillor]

A thought on the 2nd one, would one not put the whole thing under the square root aka sqt ((x^2 + 1)/x^2), bring the limit inside, and then apply lopitals rule. (Continuity on the domain), getting 2x/2x under the square root, which is still one.

[@rafaelb.333]

That's exactly what I thought

[@magranin7319]

x doesn't equal square root of x squared, that's the absolute value of x, but since the limit goes to positive infinity the absolute value of x is just x so you kind of have a point.

[@pedrosso0]

1:44 I'm new to L'Hôpital's rule, why doesn't 0/infinity and infinity/0 work as well?

[@drpeyam]

0/infinity = 0 and infinity/0 = infinity (or undefined)

[@pedrosso0]

@@drpeyam so 0/infinity is not indeterminate... huh... omg of course it is xD it always decreases

[@tgx3529]

In 2) example is the function non negative & ?strictly decreasing? (if xf(y)) the limit really exists. L=1/L. Then L=1.

[@juanpablochamorro1765]

When you start doing L'H before evaluating a possible indeterminate form, I almost had a stroke 😂. Great video btw ❤

[@creamsoda1729]

I think another good thing to talk about would be the special case that the second limit has to exist in order for the limits to be equal. e.g. lim x-> ♾ x/(x+sinx) when you use L’Hopital it’s undefined but the actual limit is just 1.

[@skylardeslypere9909]

How can you actually calculate this limit then?

[@ChucksSEADnDEAD]

@@skylardeslypere9909 I believe you need to "squeeze" sin(x) between 0 and 1, meaning that towards infinity the x will dominate the value of sin(x), eventually becoming x/x which cancels out to 1.

[@alexandersanchez9138]

6:10 Unless you've defined sin, cos in terms of complex exponential function from the get-go. Then, you don't need the angle addition formula to calculate the derivatives; so, there's no circularity.

[@nathanisbored]

if you defined them that way, have fun proving all their other properties from there as well

[@alexandersanchez9138]

They follow from the standard arguments since it's easy to verify that |cos x + isin x| = 1, |d\dx cos x isin x)| =|-sin x icos x| = 1, cos 0 + isin 0 = 1. So, (cos x, sin x) is the expected point on the unit circle; since C(+) over R is isomorphic to R^2, we're done (modulo relatively easy, if tedious, analytic proofs of Euclid's axioms).

[@NeonArtzMotionDesigns]

Ya know on the AP Calc exam, the college board always manages to screw up the spelling of L'Hopital by spelling it as L'Hospital and it's funny

[@davidgould9431]

Except Guillaume de l'Hôpital also spelled his name l'Hospital (the circumflex usually shows a missing s - the french for forest is forêt, for example, and the words are (unsurprisingly) related).

[@NeonArtzMotionDesigns]

@@davidgould9431 amazing, such Interesting stuff

[@onlytimmie]

Thank you Dr.🙏❤️

[@Songvbm]

Nice tutorial and I follow your tutorials often. BUT... I did not understand what was wrong in the last example. Would you like to explain here; if proosible!

[@noahtaul]

Hello sir, How do you prove that sin’=cos, from the very barebones basic definition of derivative?

[@eliyasne9695]

4:27 We can show it exist using the squiz theorem. (x+1)/x > √(x^2+1)/x > x/x (For positive x)

[@eliyasne9695]

@@drpeyam No, the upper bound is also valid since x+1 is bigger then √(x^2+1) for positive x ( because if you square both of them you get x^2+2x+1 in comparison to x^2+1, there's a 2x difference.)

[@quinnculver]

Is 3) really _circular_? Might it be more accurate to call it _redundant_? Seems that regardless of how the differentiability of the numerator was proven the conditions of L'H are satisfied so it can be applied.

[@IoT_]

If it had already known before doing that limit that the derivative of sin is cos, that it's redundant , but first of all it was proven that the limit of sinx/x is 1 by other methods then to find out that actually the derivative of sin is cos.

[@quinnculver]

@@IoT_ I agree. It's only circular if the limit is being used to show that d/dx(sin x)=cos x.

[@davidseed2939]

not all derivations of (sinx)’ = cos(x) involve sinx/x. eg let y= e^ix =cosx+isinx, sinx=Im(y), y’ = iy= icosx -sinx , (sinx)’ = Im(y’) = cosx. Euler's identity depends on the series definitions of sin, cos and exp. sinx/x is not used

[@ankitbasera8470]

4:34 Another way: Put x = tan X Which means X -----> π/2 Now the question becomes lim sec X / tan X = 1 X--->π/2

[@ShadowZZZ]

yeah it needs to be either 0/0 or inf/inf on both top and bottom in the limit. The proof of L'Hopitals rule however is very complicated

[@xenvector]

great video but I don't understand the last one where you say there is nothing wrong with the proof but mathematically it is wrong? I do use l'hopital going from sin x to cos x all the time and I assumed it was always correct

[@marcussuzuki4031]

I have a personal project I want to do. I am trying to derive all of the PDFS of Statistical functions. Its hard to find online. It goes over independent and identically distributed random variables. I think with your RN volume descriptions, I can start deriving the Chi Squared. After I derive chi squared, I will be working on T distribution haha.

[@PerishingTar]

I work on plowing the fields

[@sparsetable]

Ok so I didn't wait pls *ignore* 3:10 wouldn't that mean that the limit L will satisfy L = 1/L and because sqrt(x^2+1) > x L Is positive, and so L equals one? Please answer, god Peyam.

[@drpeyam]

No, that’s assuming the limit exists

[@edmundwoolliams1240]

You don’t need to use that limit no. 3 to prove the derivative of sin is cos. Because sin and cos are just defined by their Taylor series, so you can trivially show the derivative of sin is cos without using that limit (as long as you assume the power rule, which again you can prove for all integers without limit 3).

[@el4201]

Love this channel

[@Ahmed-hd4lr]

thats why you always plug in the x value

[@mooblerthomson9851]

For the second example couldn’t you just notice that the that the leading variables in the numerator and denominator are both x^1 and when the exponents of the leading variables match as x->infinity you just take the ratio of the 2 leading variables.

[@divisix024]

That’s basically the same as what he did, though.

[@vijaymohansingh9405]

Can we write like this if x approaches to infinity x^2 +1= x^2

[@ajiwibowo8736]

So how do I do the last limit without L'Hopitals rule ? :-(

[@drpeyam]

There’s a video on that :)

[@azedineabmazadine4113]

can you ad the others languages in your videos.thank you

[@_DD_15]

2 could also be done with a x=sinh(u), so you end up with a limit which is cosh(u)/sinh(u) which is cotgh(u), then the result is obvious 😱

[@dipesh-singla]

First ques can also be done by series expansion

[@neptunian6226]

If for 3 you use the Taylor Series or Im(e^ix) to define sin(x), then its not at all circular reasoning (both literally and figuratively).

[@benjaminbrat3922]

About sin(x)/x, that would be under which definition of sin and cos? As there are multiple ones, it is not necessarily circular reasoning, is it? If I define them as their Taylor series, I can show that sin' is cos without this limit, I think.

[@rogerkearns8094]

L'Hopital's rule at the moment is _Don't visit._

[@jannesl9128]

But the steps in the 2nd one arent missleading. Lets just call the limit y. We get that y=1/y => y^2=1 which implies +/- 1 as possible solutions. Since both numerator and denominator are positive only the positive answer can be correct Therefore the limit is 1

[@axelperezmachado3500]

I agree that it isn't valid to prove that sin(x)/x ---> 0 when x ----> inf using L'Hopitals rule, but once one has already proven that independently, isn't it legit to use L'Hopital in the the third example? I mean as you said there is really no logical mistake (provided that you have proven L'Hopital before independently)

[@rhc1560]

Hello Dr Peyam. Can I make a video about your second limit that is the way I found that this limit is 1 please?

[@drpeyam]

Already done, I think it’s called don’t use l’hopital

[@rhc1560]

@@drpeyam OK but can I make a video that I put on KZbin about my way to find the second limit (limit as x goes to infinity sqrt(x^2+1/x) please.

[@shyamdas6231]

I laughed really hard at the second one.

[@alexmumbere3373]

Then we can use the two formulas of Taylor innit

[@foreachepsilon]

My analysis prof preferred little o notation.

[@MrCigarro50]

Wonderful video, muchas muchas gracias. Se le agradece.

[@hopeworld8247]

Why I am getting ranchod das chanchad vibe😂 but it was a good explanation 😍

[@likeabossrofl]

Great video and in-sighting :)

[@ianluebbers5492]

Is the only circular reasoning within L'Hopital's Rule be the sinx/x? Or are there other examples? I am trying to do a proof and I have some sin amd cosine stuff happening where i use L'H

[@drpeyam]

cos-1/x is also circular reasoning

[@ianluebbers5492]

@@drpeyam would (cos^2(pi/4*((x-2)/x))-sin^2(pi/4*((x-2)/x)))/x be circular reasoning? Sry for super specific example, im just unsure how to figure out if is circular

[@frozenmoon998]

We should not always use L'Hopital's rule, true. Sometimes we need diversity, and L'Hospital's rule is a perfect candidate for that ^^.

[@Flanlaina]

I watched your ‘Don’t use FTC’ video

[@mathemitnawid]

For the last one I’ll use l’hospital anyway 😂

[@mohammadfahrurrozy8082]

On the 3rd question, Can we change the lim x->0 sinx/x to lim -> x/x?(then it become 1 too) because lim->0 sinx is equal to lim x->0 (x) ?

[@drpeyam]

😱😱😱

[@drpeyam]

No! If that were true you could replace sin(x) with x^2

[@mohammadfahrurrozy8082]

Aaah....okay dr.peyam thank you so much 😄

[@mohammadfahrurrozy8082]

@@drpeyam maybe you can make a video about this?just wondering..

[@IoT_]

@@mohammadfahrurrozy8082 there is that video

[@tomaskubicek5983]

Actually you are not right on the third one... you are assuming the way how they proved the derivatives of sin etc... that is not the only way for example you can introduce those functions axiomaticaly, with the given propertie of derivatives, and then prove that such a function exists and is unique, which gets around that circular logic

[@speedrunnertom151]

I thought he was gonna teach spanish

[@turkeypedal]

That last one never sits right with me, as it presumes both a particular proof for the derivative of sin(x), and that the student is aware of that proof. For many students, the fact that the derivative of sin(x) is cos(x) is just given to us as a definition. And, as people have pointed out in the comments before, there are apparently more advanced proofs of the derivative that do not use the fact that the limit of sin(x) as x -> 0 is 0, using Taylor expansion or other methods. It just seems to me that the person asking for the proof should explicitly specify what is not allowed to be used. That way, whether you're using something as an axiom, definition, or theorem, it doesn't matter. The alternative is that you have to prove literally everything before you use it, which is not how math is usually done.

[@ultrio325]

His videos progresses like animes that go "Yup, that's me"

[@nanigopalsaha2408]

Hey Dr πm! I used x = tan θ on the second limit and got 1. Is it valid?

[@drpeyam]

Yeah

[@noahtaul]

One I tell my students is to calculate the limit as x->infinity of (x+sin x)/x. Easy with the squeeze theorem, but it’s an infinity over infinity limit that L’Hopital doesn’t work with. The limit you get is lim x->infinity (1+cos(x))/1 which DNE.

[@vishnupriyapadhy8818]

The last part was epic👌

[@anshumanagrawal346]

Wait, doesn't this contradict L'Hopital 's rule?

[@zlatanibrahimovic8329]

@@anshumanagrawal346 No, the rule requires that the derivative limit is actually equal to something.

[@anshumanagrawal346]

@@zlatanibrahimovic8329 Yeah, I know that now but didn't at the time of posting this reply

[@epicmorphism2240]

Please can you do a talk with steve and f m? I don‘t want any negative vibes in the math community!

[@drpeyam]

???

[@epicmorphism2240]

Dr Peyam is everything ok with you and fm?

[@3manthing]

1:57 You should have said "blackboard".🤭

[@davidgould9431]

6:42 "you cannot assume what you want to show". That is, technically speaking, "begging the question". About which I believe I bored you many videos ago (you are totally forgiven for having forgotten - just keep making the videos or I really will sulk).

[@drpeyam]

I remember that, haha

[@cparks1000000]

I'm not a big fan of your example 3.

[@drpeyam]

Ok

[@theproofessayist8441]

With example #2, people have done recurrence relations with integration by parts now let's do it with L'Hopital's Rule!!! *wink* *wink*

[@chouayabdelali3241]

I never believed that maths can be funny

[@alfredocuomo4284]

Very usefull

[@bobajaj4224]

you assumed x>0 without writing it.. at - infinity it's -1

[@advaithbala]

L' Hôpitaln't

[@cjspear]

Thank you!

[@speakingsarcasm9014]

I got the second one, √(1+x²)/x Since x approaches ♾️, 1+x² is approximately x² Thus √(1+x²)/x = √(x²)/x = x/x = 1

[@rob876]

L'hopital's rule works only for f(x)/g(x) where f(x) -> 0 and g(x) -> 0 for x -> some value because f(x) ~ f(a) + (x-a)f'(a) + ... and g(x) ~ g(a) + (x-a)g'(a) + ... so in the limit, if f(a) = g(a) = 0, f(x)/g(x) -> (f(a) + (x-a)f'(a) + ...)/(g(a) + (x-a)g'(a) + ...) ~ (x-a)f'(a)/((x-a)g'(a)) = f'(a)/g'(a) This is why the first one doesn't work.

[@purim_sakamoto]

おもしろかった🙂

[@danyaaaa31]

Thank you.

[@dipesh-singla]

Dont know why are u saying rhis at1:00 but l hopital work only on 0/0 or indeterminant/indeterminant

[@drpeyam]

Someone didn’t watch the whole video 😉

[@mohammadfahrurrozy8082]

😂

[@cwwanimalsandfunforkids8726]

Nice

[@gregoriousmaths266]

For number 1 it’s not a 0/0 situation Edit: oops you said that after you did it my bad

[@otakurocklee]

No offense. But both you and blackpenredpen are propagating some bad ideas with regards to the third limit (sinx/x) and with regards to using theorems in general. And I really think this is causing a lot of unnecessary confusion among math students. Once we know l'Hopital's rule is true, and once we haven proven that the derivative of sinx is cosx, there is absolutely nothing wrong with using it in the 3rd situation. There's nothing mathematically wrong with it. Yes, we all know that to prove the derivative of sinx is cosx we need an independent way of proving that lim sinx/x = 1 without l'Hopital's rule. But that's a moot point. But once we've established that the derivative of sinx is cosx, I can use that result without concern for how it was proved. Once a theorem is established, and once we know it has been proven, it's perfectly fine to use it in all situations. The fact that the proof of the theorem being used, already has within it an independent proof of the problem you're trying to solve is completely moot. We do this in Euclidean geometry all the time. The lesson both of you seem to be teaching is that you are not allowed to use a theorem unless you know how it has been proved and its proof doesn't itself contain and independent solution to the problem. This is not how mathematics is done. The whole point of theorems is that they are tools that can be used without concern for how they were actually proved. Is l'Hopital's rule not applicable to sinx/x? Does it not satisfy the conditions established by l'Hopital's rule? Then what exactly is mathematically wrong with using it here? A lot of your viewers are getting the impression that using l'Hopital's rule in the sinx/x situation is only giving the right result by coincidence, which is not the case at all. I think this can all be avoided by simply stating, "solve this without using l'Hopital's rule" or saying to the student, "you should understand how to solve this without l'Hopital's rule". Instead you guys are saying that the use of l'Hopital's rule here is somehow invalid which is just confusing them.

[@drpeyam]

None taken, but that’s not how the education system works here. Usually here one proves the limit of sin(x)/x here using the proof with triangles, but then the students don’t care about it because they can just use l’Hopital’s rule, and they think L’Hopital’s rule is a substitute of the proof, which it is not. Of course if you can independently prove the sin(x)/x formula somehow then sure you can use l’Hopital’s, then sure, go with it, but that’s not what blackpenredpen and I are saying. We are talking about the 99% of students who do that, not about the 1% like you or me. We’re talking from experience here

[@otakurocklee]

@@drpeyam Sure... but in that case you should make a distinction between deriving theory, and applying results (as is usually the case on a test). When theorems are built, they have to be built in a certain order, and we can't assume what we want to prove it. And definitely, within that context of building the theory, the triangle proof of the limit of sinx/x is fundamental and using l'Hopital's rule to prove it in that context would be ridiculous since we haven't proven l"hopital yet. I think students can and should be taught this. That's all good. My issue is when you say it is "mathematically invalid" or saying it is "wrong" when simply calculating a limit like this. As I mentioned all this can really be avoided by just telling the student on the test, "solve this without l'Hopital's rule or solve this only using the squeeze theorem". The student will learn the triangle proof and all is good. And it doesn't give the misunderstanding that it is "mathematically wrong" to apply l'Hopital's rule in this situation. It just emphasizes that the student should know and understand the triangle proof as it is a fundamental result. I know you guys only want to improve understanding, but there's some confusion among some viewers asking... "ok, l'Hopital's rule is mathematically wrong to use here, where else can't it be used?" I'm just going by the comments I see.

[@mahrammeskini5545]

😂

[@mahrammeskini5545]

it is so simple

[@hugodiazroa]

20k views

[@vaibhavjaiswal6953]

I solved all the limits in 15 seconds without playing the video .👍☺

[@drpeyam]

But did you use l’Hopital?

[@vaibhavjaiswal6953]

@@drpeyam no

[@drpeyam]

Very good