the most interesting proof of L'Hospital's Theorem!!
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- Опубліковано 9 лют 2025
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Despite having a PhD degree in theoretical physics, the delta-epsilon notation still makes me as uneasy as some sophomore. Thanks for the neat visualization!
A PhD in electrodynamics also doesn’t help! 😅
So relatable bro
So relatable bro
Really? I find it comforting. If it appears somewhere, it's coming back to known territory in what could be a new subject. It tells me that it's not so different.
using logical symbols make it easier
(X,dᵪ), (Y,dᵧ) two metric pairs,
E⊂X, f:E➝Y,
N(z, ε)={p∈X | dᵪ(z,p)0,∃δ>0,∀x∈N'(x₀, δ)∩E:
dᵧ(f(x), y) < ϵ.
One thing to keep in mind is that nearly every implication can be rephrased in terms of quantifiers and vice versa, for instance:
x∈Ω ⇒ (P(x) true)
is the same as saying
∀x∈Ω : (P(x) true)
This realization alone boosted my analysis study, something i wish i'd taken notice much earlier
Rule of Hospital: Never fail executing a backflip! 😮
It is extremely cool to just use FTC, triangle ineq, and basic math to push the epsilon proof straight through to the antiderivative, HOWEVER it is worth noting that g' nonzero in a neighborhood of infinity is an incredibly strong condition!
LH seems to need g'(x) to be nonzero in a neighborhood of the limit point? Or how else would it work?
It's not that strong, all it means is that g is eventually monotonous
@@coc235 The limit of g being 0 doesn't imply that g is monotonous after some point
This was a real nifty one! Thank you!
L'Hospital had the s removed while he was in... well they didn't have a name for the place yet.
The letters “os” in some French words is sometimes equivalent to “ô”, so the “s” is missing in “l’Hôpital”
awesome proof,thanks for the video
The one thing I dont understand is it was never proven/shown an "M" exists s.t. for all x,t > M the absolute value of the difference between the function and the limit is less than epsilon?
We showed it was for some t > M, but we dont know M exists or is finite??
The assumption was that f'(x)/g'(x)→L. By definition, that means each ε>0 must have a choice of M to go with it.
Very nice eps delta proof for students to study.
Is the proof in Spivak book considered as standard? It only claimed that lim f'(x)/g'(x) exist and lim f(x)=g(x)=0.
I'm sure that these are not the only conditions for applying l'Hôpital's rule in Spivak's book, for the result would then be false. What we typically require in this case (0/0 indeterminate limit taken as x→∞) is:
1. f & g are differentiable from a certain point on
2. g' is never zero from a certain point on
3. lim(x→∞)f(x)=lim(x→∞)g(x)=0
4. lim(x→∞)f'(x)/g'(x) exists, or is ∞ or -∞
If these conditions are satisfied, then lim(x→∞)f(x)/g(x) exists, or is ∞ or -∞, and lim(x→∞)f(x)/g(x)=lim(x→∞)f'(x)/g'(x)
If this proof requires that g'(x) (eventually) not be zero, why don't we have to check that condition every time we want to use L'Hospital's rule?
Indeed, Michael's graph showing what the epsilon means, actually demonstrates how a function like g(x) can approach zero as required by the theorem, and yet still equal zero infinitely often, which seems to violate this ≠0 condition. What am I missing here?
@@jacemandtWhat Michael proved was not l'Hôpital's rule, but a weakened version. The standard theorem only requires that g'(x) be non-zero after a certain point, and doesn't require g'(x) to be continuous. The extra conditions were added here so that this simple proof would work.
I couldn't resist the video title.
10:01
Nice proof indeed
hospital?
Yes because this proof is a killer
@@A_Turtle_outside LOL!
The 's' is silent in this old French name. It's probably best written without the s, but with a "carrot" over the o. In American math books it is often written l'hopital, which is how it is pronounced.
The name can apparently be spelled both with or without the S. His original spelling was L’Hospital, but modern French spellings of names drop the silent S so L’Hôpital also works (with the accent over the ô).
@@TomFarrell-p9z accent circonflexe 😀😜
7:36 why do f(x) and g(x) go to 0 as x goes to infinity? They are general functions
it's one of our assumptions at the beginning
Since the L'Hopital rule is applied to the indeterminate forms 0 over 0 and the same for inf over inf , the latter can be considered with flipping the denominator and nomirator . Which yields the interval as shown and makes the functions 0 at infinity
@@yousefalharbi1134 -- * denominator and *numerator*
If they don't go to zero (nor infinity) then L f/g = L f / L g.
Le'Hopital's rule is all about handling the cases where f & g both go to zero (as here) or infinity which cannot be resolved without L'H's rule since 0/0 and inf/inf aren't in the real numbers.
The Wikipedia article on L'Hôpital's_rule does a good job of showing the necessary conditions.
Fantastic proof
And that's a good place tô top.
Well play, young man. Well played. 👏👏👏
This is the usual good points - bad points proof, where the points are all eventually good.