Real Analysis | Sequential limits in functions.

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Michael Penn

Michael Penn

Күн бұрын

Пікірлер: 38
@oussamadjeziri311
@oussamadjeziri311 3 жыл бұрын
been looking for this all day. REALLY, Thank you.
@stabbysmurf
@stabbysmurf 4 жыл бұрын
I've always wondered why we don't just use the sequential definition as the primary one. Sequences are such an effective workhorse for establishing fundamental results in real analysis, you might as well use it here as well. And it is more intuitive to say "no matter how you approach a, f approaches L" than the epsilon-delta cha-cha. I guess one reason for the usual definition is that we can define uniform convergence in a very similar way.
@Nickesponja
@Nickesponja 4 жыл бұрын
Well in order to use the sequential definition, you'd need to first define the limit of a sequence, and for that you need the "epsilon-delta cha cha"
@stabbysmurf
@stabbysmurf 4 жыл бұрын
@@Nickesponja I've found that the epsilon-N definition of a limit is a bit more obvious, while the epsilon-delta definition of continuity is a bit more opaque. It's also easier to say it plainly, e.g. "lim {xk}=L if every open interval of L contains all but a finite number of sequence elements."
@TheAAZSD
@TheAAZSD 4 жыл бұрын
@@stabbysmurf I've found that there are functions that are just easier with epsilon-delta. I think the sequential definition is really good for most functions for limits. But occasionally it just drives you into the woods for no good reason.
@backyard282
@backyard282 4 жыл бұрын
the last problem was pretty intuitive. the limit as x goes to zero of sin(1/x) means that the argument of sine tends towards infinity as x goes to zero, but limit as x goes to infinity of sin(x) obv doesnt exist due to sine just oscillating, i.e. if you picked L to be any real number you could show easily using epsilon delta that it isnt a limit.
@debendragurung3033
@debendragurung3033 4 жыл бұрын
The theorem looks neat. But again the proof just flew over my head. I think this is more concrete definition of the limit of the function.
@ashwinvishwakarma2531
@ashwinvishwakarma2531 4 жыл бұрын
Just a slight subtlety that you had to show x_n=/=a for any n. But that comes from the definition of the limit where you say |x_n - a|>0, so x_n=/=a for any n (also because x_n is chosen from A and a is a limit point of A).
@goodplacetostop2973
@goodplacetostop2973 4 жыл бұрын
19:41
@wojciechgrunwald80
@wojciechgrunwald80 4 жыл бұрын
you're so fast
@yagmur2460
@yagmur2460 3 жыл бұрын
Why you choose delta=1/n at 10.43?
@danwe6297
@danwe6297 4 жыл бұрын
Wow! Heine's theorem! I haven't seen this boi for ages! :O :D
@alijoueizadeh8477
@alijoueizadeh8477 3 жыл бұрын
Thank you.
@tomatrix7525
@tomatrix7525 3 жыл бұрын
These are unreal. Thanks michael
@PubicGore
@PubicGore 2 жыл бұрын
Actually they are real... analysis. I'll see myself out.
@Jancel705
@Jancel705 5 ай бұрын
@@PubicGore lol!
@Godél-p4e
@Godél-p4e 3 жыл бұрын
Great guy
@tylercampbell1568
@tylercampbell1568 4 жыл бұрын
What is your #MegaFavNumber ???
@사기꾼진우야내가죽여
@사기꾼진우야내가죽여 4 жыл бұрын
Since we have proved that sequential limit of function is equivalent to the original definition of limit of function, we can use this fact to show extreme value theorem.
@themathguy3149
@themathguy3149 3 жыл бұрын
you rock!
@anshumanagrawal346
@anshumanagrawal346 2 жыл бұрын
You need 0
@Jancel705
@Jancel705 5 ай бұрын
the absolute value of any number is bigger that or equal to 0 and x≠a so it's implicitly stated.
@autishd
@autishd 4 жыл бұрын
I wonder whether you could help me solve question 23 from 2020 AMC 10B. The question is as follows: Square ABCD in the coordinate plane has vertices at the points A(1,1), B(-1,1), C(-1,-1), and D(1.-1). Consider the following four transformations: - L a rotation of 90 degrees counterclockwise around the origin; - R, a rotation of 90 degrees clockwise around the origin; - H, a reflection across the x-axis; - V, a reflection across the y-axis. Each of these transformations maps the square onto itself, but the positions of the labeled vertices will change. For example, applying R and then V would send vertex A at (1,1) to (-1,-1) and would send vertex B at )-1.1) to itself. How many sequences of 20 transformations chosen from {L, R, H, V} will send all of the vertices back to their original positions? (a) 2^37 (b) 3.2^36 (c) 2^38 (d) 2^39
@wojak6793
@wojak6793 4 жыл бұрын
You know theres a solution for all AMC problems on the Art of Problem Solving website
@truonghuynhvan2998
@truonghuynhvan2998 3 жыл бұрын
The video sound is pretty good, beyond my imagination
@schweinmachtbree1013
@schweinmachtbree1013 4 жыл бұрын
definition of the limit of a function is wrong. should be 0 < |x-a| < delta instead of |x-a| < delta ...
@amandalahadi6894
@amandalahadi6894 4 жыл бұрын
That not necessary to be write. Because absolute value always greater than 0.
@schweinmachtbree1013
@schweinmachtbree1013 4 жыл бұрын
@@amandalahadi6894 |0| = 0
@amandalahadi6894
@amandalahadi6894 4 жыл бұрын
You're right. I mean in the case of limit. We don't say it limit of x to a, if we have x = a. So, |0| never happens in this case
@schweinmachtbree1013
@schweinmachtbree1013 4 жыл бұрын
@@amandalahadi6894 exactly; when talking about the limit as x->a we want to ignore the case where x=a, and so we want 0
@amandalahadi6894
@amandalahadi6894 4 жыл бұрын
Aha, I get the poin. Ok, thanks for your explanation
@franksaved3893
@franksaved3893 4 жыл бұрын
In Italian analysis courses they call it the Bridge theorem, wtf.
@alexwestworth6962
@alexwestworth6962 4 жыл бұрын
Notice how Michael structures videos similar to, and speaks using the formal language of, mathematics journal papers. And that's a good place to stop.
@orenfivel6247
@orenfivel6247 Жыл бұрын
would like to show a contra positive of ⇒ (∃{a_n} a_n→a ∧ f(a_n)↛L) ⇒ lim{f(x),x→a}≠L PF: foil out the definitions for each statemet in the "and" operation (∧) above: ∃{a_n}: ∀δ>0 ∃N(δ)∈ℕ ∀n∈ℕ, n> N(δ)⇒|a_n-a|0 ∀M∈ℕ ∃n∈ℕ, n>M ∧ |f(x)-L|≥ ε. Set M= N(δ), find n> N(δ) s.t |a_n-a|0 ∀δ>0 ∃x∈ℝ (x=a_n for some n> N(δ)), |x-a|
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