an awesome approach to otherwise mundane limits
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- Опубліковано 11 тра 2024
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For the second limit, you can work it out using the identity arctan(x) + arctan(1/x) = pi/2.
We end up with lim(x->infty) x(arctan(x) + arctan(1/x) - arctan(x)) = lim(x->infty) x * arctan(1/x).
But since arctan(1/x) ~ 1/x as x->infty, then x * arctan(1/x) ~ x/x and so the limit is 1.
Yes but the point of the video was to solve the limit in a different, creative way
@@watchnarutoshippuden3228 Yeah I know, I was just giving an alternative approach for those interested.
10:05 "a happy accident" 🙂 in memoriam Bob Ross? 😊
10:56
Pretty cool technique. using the l'hopital rule for the first one would be a mistake anyway, because you need to know this limit in order to differentiate. Well, not strictly speaking - but definitely in the usual way this is introduced. What you need is (e^x)' = e^x and I guess there are many ways to obtain this without obtaining the limit.
Well, if you apply the l'hopital rule with respect to a, then you'll pretty much get the same answer.
But you can think that this is the generalization of the typical limit (e^a-1)/a with a->0. so you know the special case and know the chain rule.
@@dinuwarabinudithdesilva5464but given limit is derivative of x^a since when a --> 0, (x^a)' = limit of (x^a - x^0)/(a - 0) = limit of (x^a - 1)/a
So we "can't" use l'Hôpital since it will lead to circular reasoning
@@El0melette -- * *chain*
@@robertveith6383 xd sorry, non-native speaker here.
That's a clever trick! I need to learn more about dominated convergence...
really useful, I learnt it in first year after high school, first year of bachelor
I used this at Uni in the 1980s. I don't remember every using L'Hopital's rule.
Conversely, the first example reminded me of a solution in the SE question: Demystify integration of ∫1/x.dx
Incredible!
This is an very interesting theorem, and reminds me of the Weierstrass M test for series that ive learned recently. There are any connection between them?
The not creative way is to say
lim_(a->0) (x^a - 1)/a
= lim_(h->0) (x^(0+h) - x^0)/h
= (d/dy x^y) (y=0) = (ln(x) x^y) (y=0) = ln(x)
Okay - I will try to describe a function. I claim it infinitesimally fills complex plain but it is only continuous along single line and non-continuous everywhere else. I suppose it can be extended to ℝ² and nestings of n-spaces (don't know enough about non-finite spaces)
let x∈ℝ, altho it could be [0, ∞] define z=x(cosx + isinx) then z fills complex plain but spoils continuity reducing it to case where continuity is fixed for a single infinitesimal function zₓ where zₓ = x(cosx +isinx) for single infinitesimal x in ℝ
Reasoning every zₓ or its equivalents in ℝ², is limited to continuity along zₓ
Interim conclusion: thus z partitions complex plain or ℝ² or any nestings of ℝ² as a squished spiral in a plain - or projected spiral in n-space onto any n-plane or nestings of n-plains where n is a variable will make for an infinitesimally dense thing while spoiling cntinuity off zₓ
Apologies if this makes little sense.
Taking z on [-∞, ∞] seems to apply a pleasant symmetry helpful to physicists? Conic sections?
Even where there are overlaps due to angular behavior on [0, 2π] each infinitessimal line has different acceleration and velocity
You're using 'infinitesimal' or the associated adverb in a rather strange way. I think 'discrete' or 'singular' is the adjective that better suits your description, in some cases.
You define zₓ to be a point in the complex plane yet you call it an infinitesimal function, whatever that is. Also, z is both complex and on the real interval (-∞, ∞). Lastly, in the absence of a time coordinate, you don't have velocity and acceleration but rather first and second derivatives (temporal vs spatial).
To be clear: I am not saying what you tried to describe makes no sense, simply that I couldn't make sense of it. It may SEEM that infinitely many Archimedean spirals will tessellate ℝ² or the complex plane but, in fact, they do not... or maybe that part isn't important to your construction, I'm not sure.
@@physicsjeff now thinking about nested functions something like h(f(f(x))) but it is making my head hurt
EDIT: h(x, θ) = x (sinθ + i sinθ)?
Those are simultaneous equations...somehow.
X=1/a, a=r+x,
Answers are ln(x) and 1
so don't you plug in ln(1) and zero back?
Limits
Gotta love using first year grad school techniques to do high school calc limits rofl
measure theory is second year undergrad in the UK
@@padraighill4558 Damn, second year undergrad in the US is usually series and multivariable calc and linear algebra. At UCLA where I went there was sometimes a third quarter analysis course 131c offered to undergrads that did measure theory (first two quarters were Baby Rudin) but I think it might have been only offered once a year. Otherwise you'd learn measure theory and the Lebesgue integral in 245a which was a grad course.
@@homerthompson416 Oh wow, I'm currently a second year undergrad at Oxford and we did all this stuff (Measures, Defining an integral, Lp spaces) however I much prefer Algebra (Rings, Modules) and Topology (Analytic, Algebraic) all of which we did this year.
@@padraighill4558 Come on, everyone's going to look bad compared to Oxford haha. Algebra was definitely my favorite sequence in undergrad but if you ever told anyone damn my algebra class is killing me they'd look at you like you were a moron thinking you mean quadratic formula and FOIL and talk about how easy algebra was. Then it was always fun to pull out my copy of Lang and ask for help with a problem.
@@homerthompson416 Yess I agree. My favourite for that is Number Theory. The questions are so incredibly easy to understand (primary school) but the solutions are deceptively tricky!
too esoteric for me
Is the first limit not circular logic? Wouldn't we have needed to know that limit to show that the integral of 1/t is the natural log of t, which you used in the calculation?
There are lots of ways of defining the natural log of t that produce the result that its antiderivative is the reciprocal of t with needing any recourse to limits. For example define natural log such that if x = ln(y) then y = e^(x), where e^x is the function whose derivative is itself. Then dy/dx = y and hence dx/dy = 1/y.
@@RexxSchneider But in your logic the fact that there is a number e such that the derivative of e^x is itself is tantamount to showing that the desired limit exists and there is a number x for which it is equal to 1. I guess you could linearly order the work by using the first half of the proof (using Lebesgue's dominated Convergence theorem to show the limit exists and equals the integral of (1/t)dt from t=1 to t=x) to show the limit exists for all positive x and that there is a number e so that if x=e the limit is 1, (perhaps using the intermediate value theorem and some lower bounds on the integral of 1/t), then using that fact to show that the derivative of e^x is itself and therefore that the natural logarithm is the antiderivative of 1/t, which allows you to show the final step.
@@TheElihs It is possible to define ln(x) as that integral and then prove that its inverse is an exponential. I’ve heard that that how it was originally defined.
@@TheElihs You could do all of that, but you could also simply define exp(x) as the function whose derivative is itself, and ln(x) as its inverse function. I simply don't agree that you _need_ to start with a limit of an expression like (1 + 1/a)^ax. You can derive either definition from the other, and there is nothing that forces you to use a limit definition.
The steps in this video are completely self-contained without any need for a circular reference.
Is this the (in)famous "dominated convergence theorem"??? Someone pinch me please XD
आलु खाऊ, रन्दी!! म तपाईंको ठुलो फलोअर हुं।
Wow, almost first
This is pointless - not "creative".
Michael I love you and your videos, but your haircut here is HORRIBLE, it is so horrible that it’s distracting me! 😂😂😂😂😂
Please, change your barber or whatever NOW! ❤❤❤
his haircut here is GREAT , YOUR distracting us, be quiet !
@@charleyhoward4594 -- You're = You are