L’hopital reminds me of my classmates in AP physics, they copied off of me for everything yet since I didn’t do much homework they had better grades than I did.

What a video, I knew the proof but I never imagined the scandal. thank you.

This is a great video! Love the different perspectives to the explanation

L'hôpital taking credit for his students work is a lot like the people who took credit for my biophysics equations and never cared for acknowledging me

@subhasish-m

4 жыл бұрын

That's unfortunate :( Could you tell us a bit more?

So, actually... I neglected or ignored to mention in my previous comment that the contract between mr l'Hopital and Johann Bernoulli, although it has the appearance of a legal agreement, and mr l'Hopital acknowledged Bernoulli and Leibnitz in his book, the contract is still a slave contract... In addition others have commented here that they have been taken advantage of in their scientific work, and so it is important to mention this as well. In addition according to Wikipedia en.m.wikipedia.org/wiki/Johann_Bernoulli Johann Bernoulli appears to have been quite evil, being jealous of his older brothers Jacobs teaching position and even stealing from him and his own son Daniel. So, it appears that due to this Johann Bernoulli was quite greedy in his desire for a highly esteemed teaching position, and mr l'Hopital took advantage of that, and bound mr Bernoulli with his slave contract... to which Johann Bernoulli apparently consented, and sold himself like a slave to mr l'Hopital, although he has been taking advantage all by himself of his older brother Jacob and his son Daniel. So it is what it is with this story... Still, these kinds of slave contracts do exist, even today. For example, when I started as a research assistant at the Royal Philips research laboratory, Eindhoven, Netherlands, there was this dreaded article 6 in my contract, that stated kind of the same thing, namely, what ever I may invent in this employment, is company owned, and even after leaving I was not allowed to practice similar activities elsewhere... And when I asked whether this is legal, they told me, "that this is not really being used in this way", but the mystery remains that if it is not really being used like that, why is it in the contract in the first place? And the answer is, that these kinds of contracts are not exactly legal, and so the stronger party such as a company or a government is being hypocritical about this, but being the stronger party they can always abuse your rights, any way they please but... and here it comes... ready? ... as long as they can get away with it, under the false appearance of a legal agreement. And this attitude looks very similar to what mr l'Hopital wrote in his book acknowledging mr Bernoulli and Leibnitz, stating that they deserve the credit they want, so nulling out his legal agreement with mr Bernoulli, but continuing, that he is pleased with everything they leave him, so those ideas he can use and get away with it.... Finally, I have been checking up a little bit on Dr Peyam, and I have noticed that despite his great mathematical talents, apparently Dr Peyam has not been able to grow past his function of assistant professor, at two universities, the keyword here is "assistant", and so, taking everything into account, about this taking advantage of others attitude with slave contracts or otherwise, this kind of worries me... That said, once you may realize that you are possibly being taken advantage of, you are not supposed to become evil as well like apparently Johann Bernoulli, and so this is important as well.

Thanks for the wonderful illustrations with the 'racing cars' and tangent-slope images; very helpful!

This make so much sense and meaning

I love your channel, youre very amazing

Ultimate explanation

L'hopitals rule was a life saver while solving indeterminate limits 😂

This is a wonderful follow-up to your previous video showing the original limit the man wanted to solve. But I think only a small percent of people who know the rule can justify it intuitively, and the proof wasn’t covered in my Analysis class, but it is proved in Stewart’s Early Transcendentals in the appendix

What a great explanation!

As you drew it, the cars start at very high speed and decelerate.

if at 0 both function are at zero... then, by the definiton of the derrivative : at dx they are at f(0) + f'(0)*dx = 0 +f'(0)*dx =f'(0)*dx . Since we are divinding f(x)/g(x), the dx cancels out and if you do the same thing for g(x), you get [ f ' (x)/g' (x) ]. Since the function is continuous (Otherwise you couldn't have derrived it, f(dx)/g(dx) = f(0)/g(0).

@poutineausyropderable7108

4 жыл бұрын

@@papa15891 Its an easy proof for intuitions sake

My intuition of L'H was that it compared Taylor series expansions at a. This helped a lot when I needed to match asymptotic expansions in ODE's, PDE's, and IE's.

In Matematic Forum I have seen the example lim(x to x to x to x to x - x to x to x to x)/(1-x)^5 for x go to 1. Classical calculation I lasted only until (x to x to x - x to x)/(1-x)^3. Then I saw the "probable result".😊

Wonderful

I watched this video hoping that you may say something about the this rule name because I read about it and I needed someone to confirm . Thank You very much

Great video as usual

What about the infinity over infinity form? How would we go about doing an informal proof of L' Hopital's rule working there?

@drpeyam

4 жыл бұрын

Oh, just apply l’Hopital to f(1/x)/g(1/x)

@drpeyam

4 жыл бұрын

Because x goes to infinity, 1/x goes to 0

@Green_Eclipse

4 жыл бұрын

@@drpeyam I apologize. I don't understand. Given that when x->a then f(x)->infinity and g(x)->infinity, how could we turn that into f(1/x)? We could say lim x->a f(x)/g(x) =lim x->a [1/g(x)]/[1/f(x)] which gives us the 0/0 form But using L'Hopital just makes it more complicated.

@drpeyam

4 жыл бұрын

Oh, sorry, I thought you meant lim x goes to infinity of f/g

@drpeyam

4 жыл бұрын

For infinity over infinity, yeah, I’d use the the (1/g)/(1/f) trick and multiply top and bottom by x-a to get 1/g’ over 1/f’ which gives f’ / g’

But basically.. the equation would work right to left also, right? If you integrate f(x) and g(x) instead of derivating it would work also, even usually derivating is easier than integrating (except maybe for ln() and 1/x -function types)

Isn't it enough to just pluck in the Taylorexpansion for f and g, then cancel (x-a) or why doesn't this work? i.e. lim{x->a}(f(x)/g(x))~lim{x->a}([f(a)+f'(a)(x-a)+f''(a)(x-a)^2+...]/[g(a)+g'(a)(x-a)+g''(x-a)+...])=lim{x->a}([f'(a)+f''(a)(x-a)+...]/[g'(a)+g''(a)(x-a)+...])=lim{x->a}(f'(a)+f''(a)(x-a)+...)/lim{x->a}(g'(a)+g''(a)(x-a)+...)=f'(a)/g'(a) if it exists

@drpeyam

4 жыл бұрын

I feel that might be circular reasoning, but not sure

@martinepstein9826

3 жыл бұрын

This is my preferred approach but you have to be careful. All we know is f and g are once-differentiable at c. To say they're equal to a power series centered at c is a much stronger assumption. What you want to do is write f(c + h) = f(c) + f'(c)h + o(h) g(c + h) = g(c) + g'(c)h + o(h) The 'little o' notation o(h) stands for some function satisfying lim_(h -> 0) o(h)/h = 0. (Warning: different occurrences of the symbol o(h) may stand for different functions). If f and g are differentiable at c then the above equations are true by definition. So if f(c) = g(c) = 0 and g'(c) =/= 0 we have lim_(h -> 0) f(c + h)/g(c + h) = lim_(h -> 0) (f'(c)h + o(h))/(g'(c)h + o(h)) = lim_(h -> 0) (f'(c) + o(h)/h)/(g'(c) + o(h)/h) = f'(c)/g'(c)

Ok now why did this not come in my notifications?

@drpeyam

4 жыл бұрын

It’s unlisted

@nanigopalsaha2408

4 жыл бұрын

@@drpeyam But why?

@abeb.7578

4 жыл бұрын

@@nanigopalsaha2408 yeah have the same question

Sir you said this proof is not rigorous......but I dont understand why it is not rigorous....can you plz explain.....thank you

@martinepstein9826

3 жыл бұрын

I think the only drawback of this proof is it doesn't handle the case when (x -> a) f'(x)/g'(x) is also an indeterminate form. Otherwise it's perfectly rigorous AFAICT.

Isn't direct with Taylor development?

Yes sir.. but actually i think that l'hopital 's rule says that if you are in the form 0/0 or inf/inf, and the lim f'(x)/g'(x) EXISTS, than the lim f(x)/g(x)= lim f'(x)/g'(x)... For example lim as x goes to inf of (ln(x)-cos(x))/x. If we use l'hopital rule, we'll end up with 1/x+sinx which limit does not exists... However It is easy to see that this limit goes to 0, bit l'hopital fails..

@drpeyam

4 жыл бұрын

Yes

Excuse me, Sir. What do you mean at the beginning, "a limit of infinity over infinity or zero over zero"?

@drpeyam

4 жыл бұрын

Infinity Minus Infinity kzread.info/dash/bejne/m4t4ysdqfLfcZZc.html

Why is this unlisted?

sir can you suggest any Advanced Calculus textbooks for Advanced learners please...

@drpeyam

4 жыл бұрын

Read an Analysis book like the one by Pugh or the one by Fitzpatrick, or read Apostol

Nice video ;)

Is it L'hospital or L'hopital?

@drpeyam

4 жыл бұрын

Both are acceptable, Hôpital or Hospital

What's up with all those week old comments?

@drpeyam

4 жыл бұрын

It was unlisted

Where is bprp????

@drpeyam

4 жыл бұрын

He’s fine

@gourabghosh5574

4 жыл бұрын

@@drpeyam why is not he uploading video???

@drpeyam

4 жыл бұрын

Just taking a break

@gourabghosh5574

4 жыл бұрын

@@drpeyam I like you two so much.

@gourabghosh5574

4 жыл бұрын

@@drpeyam and please make a complete playlist on contour integration from the beginning. I have only class 12 knowledge in mathematics.

I just came to heat the pronunciation 'L' Hôpital' most of the people pronounce it as 'hospital'

@jadegrace1312

4 жыл бұрын

How? There's not even an s in it