I am very glad this video exists, since L'Hôpital's theorem is one of the most misunderstood theorems, and this is not helped by the fact that in an educational setting, the idea of indeterminate forms always becomes introduced, even though the concept is essentially obsolete. The healthier way to understand the utility of L'Hôpital's rule is not by looking at indeterminate forms, but by understanding the limit theorems. In particular, if lim g(x) (x -> x0) = 0, then there exists no real number L such that lim 1/g(x) (x -> x0) = L. As such, we cannot use the limit theorems to say that lim f(x)/g(x) (x -> x0) = lim f(x) (x -> x0)·lim 1/g(x) (x -> x0). _This_ is where L'Hôpital's theorem comes into play: this gap in the algebraic limit theorems is essentially "filled in" by L'Hôpital's theorem.
@jiahuasong7382
8 ай бұрын
In your last example, you made a circular argument. Doing L'Hospital for lim sinx/x is assuming derivative of sinx is cosx. However, deriving derivative of sinx uses lim sinx/x to get cosx.
@brightsideofmaths
8 ай бұрын
L'Hospital is not the only way to calculate limits ;)
@thea.igamer3958
2 жыл бұрын
Would be great if you could make a series on non standard analysis, infinitesimals construction and Infinitesimal extensions to their standard counterparts (standard theorems). Hope L'Hopitals rule for all extensions, infinity/infinity, as x approaches infinite, 0/0 etc are coming as well !.
@rob5467
2 жыл бұрын
At a later stage could you remark on other calculus’s, for example Ito or Malliavin calculus’s and what they share or do not share of the theoretical underpinnings that you have so expertly guided us through in this series on Real Analysis?
@Acssiohm
2 жыл бұрын
At 6:56 you said two times the same thing ( the interval [xn ; x0] ) instead of adding the case [x0; xn] Beside that, I really apreciate this analysis playlist
@brightsideofmaths
2 жыл бұрын
Thanks! I hope that the little mistake was not confusing.
@Acssiohm
2 жыл бұрын
@@brightsideofmaths Not for me, I barely noticed it
@BedrockBlocker
2 жыл бұрын
I have a (probably wrong) idea of how to cover the improper case x-> +∞ (-∞ is analogous): Does it not hold that lim x->+∞ f(x)/g(x) = lim x->0+ f(1/x)/g(1/x)
@brightsideofmaths
2 жыл бұрын
Yes, this goes into the correct direction. However, this is still not the case I covered here. In the next video, I will tell everything about the other cases :)
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