limit of a sequence of continuous functions, need not to be continuous

  Рет қаралды 22,234

blackpenredpen

blackpenredpen

Күн бұрын

Пікірлер: 151
@rubensenouf1813
@rubensenouf1813 6 жыл бұрын
Thank you for all of your work ! You're an amazing teacher and you make me like and understand math even more !
@General12th
@General12th 6 жыл бұрын
I thought you were going to talk about the _product_ of all those functions, IE: f(x) = product(fn) = product(x^n). But just the limiting function is cool too.
@Koisheep
@Koisheep 6 жыл бұрын
I got functional analysis vibes from this, when you learn about L²([a,b]) and work with the metric
@rudboy9599
@rudboy9599 6 жыл бұрын
Konhat Lee Sakurai yes! Loved that class. Uniform and pointwise convergence. L^1,2,p spaces. Fun times
@volcanic3104
@volcanic3104 6 жыл бұрын
f(-1) is in a superposition!
@danielbenyair300
@danielbenyair300 6 жыл бұрын
VOLC4NICxSP4RK Lol...
@danielbenyair300
@danielbenyair300 6 жыл бұрын
VOLC4NICxSP4RK It is not both!... it is nitere!!
@waishingtseung6930
@waishingtseung6930 4 жыл бұрын
✓✓✓✓✓✓✓
@frankielai5753
@frankielai5753 6 жыл бұрын
Lol....colour consistency.....love it.
@ffggddss
@ffggddss 6 жыл бұрын
Nicely explained! ∞ can also ruin continuity when you add continuous functions. Ex.: a square wave (with averaged value assigned at each of its discontinuities) is a discontinuous function. Yet it is equal to its Fourier series, in which each term is a continuous function. Fred
@rudboy9599
@rudboy9599 6 жыл бұрын
Hence why uniform convergence is so nice
@thexoxob9448
@thexoxob9448 6 ай бұрын
in general: f(x) = limn->inf x^n is ambiguous because it doesn't specify by integers or real numbers. You put x belongs to what set below the arrow. If you are curious if n is integer it's in the video but if real numbers the negative part is gone
@JayTemple
@JayTemple 2 жыл бұрын
When I took Functional Analysis, this was literally the first example they showed us of a sequence of continuous functions whose limit was a non-continuous function.
@splodinatekabloominate846
@splodinatekabloominate846 6 жыл бұрын
math 🅱roke
@patrickgleason3053
@patrickgleason3053 6 жыл бұрын
Infinitely many infinites #YAY
@SteamPunkLV
@SteamPunkLV 6 жыл бұрын
:P
@h3bb1
@h3bb1 6 жыл бұрын
Would it be correct to say that this infinity function is continous at -1
@ffggddss
@ffggddss 6 жыл бұрын
Yes! Remember, the definition: f(x) is continuous at x=a iff lim[x→a⁻]f(x) = lim[x→a⁺]f(x) = f(a) This is true for all a in the open interval (-1,1). Fred
@h3bb1
@h3bb1 6 жыл бұрын
ffggddss awesome and thank you for the reply
@assiddiq7360
@assiddiq7360 3 жыл бұрын
Visually, the graph would converge to a reflected L-shape; a horizontal line from (-1,0) to (1,0), and a vertical line from (1,0) towards (1,∞). The beauty of the graph is that it visualize how 1^∞ is indeterminate, although only for positive number
@HeliosBeats
@HeliosBeats 6 жыл бұрын
it's the opposite for x^4, the higher n is n f(x)= x^n, (when it's even), the thinner/taller the graph
@ruhanbhandohal9876
@ruhanbhandohal9876 6 жыл бұрын
ReDesign Beats He is referring to x^n with the domain from -1 to 1. In that case it gets wider.
@HeliosBeats
@HeliosBeats 6 жыл бұрын
oh true
@pablojulianjimenezcano4362
@pablojulianjimenezcano4362 6 жыл бұрын
The t-shirt is incredible, best idea ever! #tshirt #Yay
@twwc960
@twwc960 6 жыл бұрын
It's also the case that infinite sums of continuous functions need not be continuous, even if the infinite sums converge everywhere: consider the Fourier series for the square wave function, for example. Each term in the series is continuous, and the series converges everywhere on the real line, but the function it converges to is not continuous. In fact, before Fourier, most mathematicians believed that series of continuous functions, when they converged, converged to continuous functions. Fourier found the first counterexamples, but his work was not very rigorous. I think Dirichlet was the first to establish these results rigorously.
@Guil118
@Guil118 6 жыл бұрын
#t-shirt Man! Never have I ever considered buying any merch from any youtuber. But yours made me change my mind
@Guil118
@Guil118 6 жыл бұрын
Also Im tackling the 3n+1 problem and its driving me insane!
@jorgesponja3042
@jorgesponja3042 6 жыл бұрын
#t-shirt hype!! Your videos are preparing me for starting college in 2 weeks, keep doing it, you're the best #BPRP
@ogorangeduck
@ogorangeduck 6 жыл бұрын
不错! also that T-shirt is LIT
@blackpenredpen
@blackpenredpen 6 жыл бұрын
orange duck 謝啦!
@hopp2184
@hopp2184 6 жыл бұрын
Do a video about tetrations!
@blackpenredpen
@blackpenredpen 6 жыл бұрын
I will, one day.
@mike4ty4
@mike4ty4 6 жыл бұрын
And what happens for negative values of n - i.e. "f(x) = x^{-\infty}"? You get the opposite effect - everything outside of [-1, 1] is okay, but the divergent bit is inside!
@Shsbzyqn123
@Shsbzyqn123 Жыл бұрын
Good going sir
@SeriousApache
@SeriousApache 6 жыл бұрын
Calculating around infinity it's like approach speed of light, normal law stop working here.
@General12th
@General12th 6 жыл бұрын
Laws work just fine as we approach light speed. Special relativity handles it all perfectly well.
@GarryDumblowski
@GarryDumblowski 6 жыл бұрын
It does, but I think they were referring to Newtonian physics. The strange thing is, due to the boundlessness of infinity, ordinary numbers all have the same properties. Even huge numbers like TREE(3) or G(64) have all the same properties as other numbers; they have prime factorizations and you can take their square roots and graph functions involving them. Same goes with this; the graph of x raised to some ginormous power is still perfectly continuous. Putting it in the context of a limit almost makes this property seem to instantaneously jump to unruliness, so it's quite separated from the more gradual nature of relativity
@SeriousApache
@SeriousApache 6 жыл бұрын
So it is no longer normal laws
@General12th
@General12th 6 жыл бұрын
RUSapache There's nothing "normal" about Newtonian physics.
@GarryDumblowski
@GarryDumblowski 6 жыл бұрын
I mean, that's a fair assessment, but it's "normal" in the sense that at the pitiful speeds we humans are able to achieve, you have to know where to look to see the effects of relativity. It's what's considered "normal" to us.
@yrcmurthy8323
@yrcmurthy8323 6 жыл бұрын
#YAY Hats off to you !
@chupetaparabose1
@chupetaparabose1 6 жыл бұрын
I have a question about the solid numbers , for example i have the limit of 0/(x*x) with x-->Infinity. By me it should be 0 , but i m not really sure about it...THX
@samokoribanic9184
@samokoribanic9184 6 жыл бұрын
#YAY
@muse0622
@muse0622 4 жыл бұрын
How 'bout differentiating the x^inf?
@estuardodiaz2720
@estuardodiaz2720 6 жыл бұрын
Hey, but what about x being equal to 1+1/n. By def. of e, lim n-> infinity (1+1/n)^n = e, so, in the limiting function, you would have more than one point different from 1 converging to a number different than 0, right??
@thomashaas4381
@thomashaas4381 6 жыл бұрын
No, because the limit function f(x) = lim {n->inf} f_n(x) is independent of n. By definition of f, you can't have the argument x be dependent on n. You basically pick a fixed value for x and then look at the limit of the sequence f_1(x), f_2(x), f_3(x), ... which you then define to be f(x). You see that x is fixed for the sequence.
@estuardodiaz2720
@estuardodiaz2720 6 жыл бұрын
oh, okay, but what if we fix x to be lim {k->inf} (1+1/k) or perhaps x = 1+ epsilon, I know in this case we cannot use the definition of e, but what assures us that any x grater than 1 diverges, perhaps the boundry of 1 does not (thanks for answering by the way haha)
@thomashaas4381
@thomashaas4381 6 жыл бұрын
Well if you fix x = lim {k -> inf} (1 + 1/k), then x = 1. Hence you again get f(x) = 1 as result. x = 1 + epsilon > 1 (assuming epsilon > 0), hence you get divergence. The only way to get what you want is to look at the sequence f_n(x_n) where x_n = 1 + 1/n. However, that is not how the limit function was defined because the argument x has to be fixed/constant (it can't be sequence that is dependent on n)
@estuardodiaz2720
@estuardodiaz2720 6 жыл бұрын
oh, okay, I think I got it, thank you!! :)
@n0ita
@n0ita 6 жыл бұрын
Pretty cool, i like to imagine this fuction as a inverse black hole, everything inside stay safe, but outside, it becomes bizarre...
@blue_blue-1
@blue_blue-1 6 жыл бұрын
That‘s why greek philosophers didn‘t dare to think about infinity. Healthy philosophy.
@Bhamilton-ws4go
@Bhamilton-ws4go 6 жыл бұрын
Funny thing to point out, but at the beginning, x⁴ is a skinnier parabola, not a fatter one, than x² ;)
@blackpenredpen
@blackpenredpen 6 жыл бұрын
the bottom part... : )
@Bhamilton-ws4go
@Bhamilton-ws4go 6 жыл бұрын
blackpenredpen I guess with parabolas, it's all about perspective, since every parabola is similar :)
@KnakuanaRka
@KnakuanaRka 6 жыл бұрын
Well, actually the part of the curve below 1 is fatter, but the rest above is skinnier. With the scale of his drawings, they stop around 1, so they’re perfectly right.
@factsheet4930
@factsheet4930 6 жыл бұрын
I feel like I need to point out that while the function is 0 between -1 and 1, it doesn't necessarily jumps to 1 at x=1... since... x is approaching 1 it's like saying: lim x->1 of x^infinity... which is undefined but personally ill just continue the curve upwards to 1 at very close to 1 instead of jumping because the values should map there :>
@stephenbeck7222
@stephenbeck7222 6 жыл бұрын
No think about the “r” example (like geometric series). The left hand limit as x goes to 1 is indeed 0. The right hand limit is infinity. And the value of f(1) itself is 1. A very unusual graph but that is what makes it a great example.
@ffggddss
@ffggddss 6 жыл бұрын
For each finite value of n, the curve goes smoothly up to 1 at x=1. In the limiting case, it actually *does* jump from 0 to 1; it's discontinuous there. Fred
@harrisidh
@harrisidh 6 жыл бұрын
okay but how about we look at it as lim (1-1/n)^n = 1/e n→∞ so it'd have to pass through 1/e when it's very close to 1
@factsheet4930
@factsheet4930 6 жыл бұрын
But that's a different limit question sadly... look at the graph of: y=(x+1)^(1/x) this graph does have a point which 1^infinity appears (when x=0) and yet it doesn't pass through 1/e! (in fact it never has a value of 1/e because as x->infinity, y->1). The reason I believe it does go up to 1 as lim x->1, is because for any other value of n (in x^n) it does go up to 1...
@thomashaas4381
@thomashaas4381 6 жыл бұрын
Just to make it clear: the limiting process in Blackpenredpen's case is with respect to n. The limit function f(x) is the limit of the sequence f_1(x), f_2(x), .... In case of x = 1, the sequence is constantly 1 and hence the limit is 1 as well. At no point does he consider a limit with respect to x. But since lim x->1 f(x) = 0 != 1 = f(lim x ->1 x) = f(1) we see that the function is discontinous at x = 1.
@adamzeggai5506
@adamzeggai5506 4 жыл бұрын
the graph of even numbers doesn't get wider, it actually gets tighter the greater the value of n gets lol
@andresxj1
@andresxj1 6 жыл бұрын
But the divergence in the negative part (x1), isn't it?
@sikandarbakht2076
@sikandarbakht2076 6 жыл бұрын
What about t-shirts, i cannot wait much, what have u thought about it??....BTW great video... Thanks
@blackpenredpen
@blackpenredpen 6 жыл бұрын
Sikandar Bakht it should be coming soon.
@MrYadaization
@MrYadaization 6 жыл бұрын
Could you do a video on quaternions?
@thexoxob9448
@thexoxob9448 6 ай бұрын
In general: the limit of a sequence may not have the same properties as the sequence, like 1.4, 1.41, 1.414, 1.4142, .... These numbers are rational, but if you take the limit as the sequence goes to infinity, the limit is irrational. (Take a guess what number it is)
@jercki72
@jercki72 6 жыл бұрын
lol I saw the thumbnail and knew exactly what it was about
@weykasemolibdenodio7184
@weykasemolibdenodio7184 6 жыл бұрын
Functional Analysis!
@milos_radovanovic
@milos_radovanovic 6 жыл бұрын
Great video! And this time finaly no contraversal statments! You are geting more consistant! Will you make any suplemental for that video you puled down on infinite nested square roots?
@milos_radovanovic
@milos_radovanovic 6 жыл бұрын
I also think that videos on integral and derivative of infinite nested square roots deserve some added rigor on domain definitions!
@blackpenredpen
@blackpenredpen 6 жыл бұрын
This is actually a prep for that : )
@milos_radovanovic
@milos_radovanovic 6 жыл бұрын
Just keep a clickbait at bay, and all will be well! ;D
@ReyhanMehta
@ReyhanMehta 6 жыл бұрын
But is infinity an integer?
@revilo7389
@revilo7389 6 жыл бұрын
What about at x = (1 - 1/n)?
@thomashaas4381
@thomashaas4381 6 жыл бұрын
the limit function has no parameter n hence you can't really look at f(1-1/n) unless you assume n to just be some constant value in which case you have f(1-1/n) = 0 since 0
@JamalAhmadMalik
@JamalAhmadMalik 6 жыл бұрын
#yay!
@EMorgensztern
@EMorgensztern 6 жыл бұрын
can you find the continuity (or not) of y=x^(1/x) from -inf to 0 ? I love your videos about complex # Thanks for what you do #yay from France !
@helloitsme7553
@helloitsme7553 6 жыл бұрын
Eliott Morgensztern x cannot be 0 since 1/0 is not possible. Ehm x cannot be -2 either , cause roots of negative numbers. I think x cannot be in the form of -2a where a is a natural number, but idk how to prove there are no other incontinuities
@Vojtaniz01
@Vojtaniz01 6 жыл бұрын
HelloItsMe That is enough to prove that the function y=x^(1/x) is not continuous on given interval. No need to search for any other discontinuities.
@ahmadkalaoun3473
@ahmadkalaoun3473 6 жыл бұрын
You said 1 to infinity's powee is equal to 1...but i don't think so... We have Lim(n->infinity) (1+1/n)^n=e If we remplace n by infinity we get (1+0)^inf=e 1^inf=e and - 1^inf=-e What do you think 🤔 🤔🤔
@ffggddss
@ffggddss 6 жыл бұрын
But you can't use that argument to define 1^∞, because lim[n→∞] (1 + a/n)ⁿ = eª which, if you "set" n=∞, says (1+0)^∞ = 1^∞ = eª So you can get any positive number you want out of that. Fred
@ahmadkalaoun3473
@ahmadkalaoun3473 6 жыл бұрын
ffggddss you're right... So we can say 1^inf=0/0=inf/inf Al of those terms are indefinite... Aren't they?
@MiguelGonzalez-hy4sd
@MiguelGonzalez-hy4sd 6 жыл бұрын
7:51 that wouldn't be a indeterminate?
@icew0lf98
@icew0lf98 6 жыл бұрын
would arctan(f(x)) be pi/2?
@L1N3R1D3R
@L1N3R1D3R 6 жыл бұрын
Well, f(x) is a piecewise function, so any composition of that function would also have to be defined piecewise as well. For x 1, f(x) always tends to positive infinity, so arctan(f(x)) would indeed tend to π/2; however, I'm not confident enough with defining values for infinity as opposed to just saying the limiting value, so think of that how you will.
@blackpenredpen
@blackpenredpen 6 жыл бұрын
Depending on the domain
@icew0lf98
@icew0lf98 6 жыл бұрын
yeah I forgot to mention "for x larger than 1"
@xaxuser5033
@xaxuser5033 6 жыл бұрын
#Yaaay
@alexdarcovich9349
@alexdarcovich9349 6 жыл бұрын
Cheeky :) #yay
@helloitsme7553
@helloitsme7553 6 жыл бұрын
I thought you were gonna talk about 1+x+x^2+x^3..... which is not continuous at 1 cause it's 1+1+1+1+1+1+1+1 and also 1/1-x = 1+x+x^2... If you plug in x you get 1/0
@blackpenredpen
@blackpenredpen 6 жыл бұрын
Oh yea, that too!
@NAMEhzj
@NAMEhzj 6 жыл бұрын
but that limit (1 + x + x² + x³ + ...) is actually (kinda) fine. It just diverges for |x| >= 1, but if you take all of the domain, namely the open interval (-1, 1) its just 1/(1-x) which is perfectly continuous there.
@helloitsme7553
@helloitsme7553 6 жыл бұрын
NAMEhzj yeah I know , only discontinuous on |x|>1 or equal to 1
@NAMEhzj
@NAMEhzj 6 жыл бұрын
i just meant that the problem at 1 is not one of continuity, it just isnt defined there.
@hankseda
@hankseda 6 жыл бұрын
Hard to be rigorous on pointwise convergence in less than 12 minutes. 👏
@blackpenredpen
@blackpenredpen 6 жыл бұрын
Yea... I just wanted to provide an example in this video
@willbishop1355
@willbishop1355 6 жыл бұрын
Doesn't really make sense to write down 1^inf = 1, as you've pointed out in other videos, e.g. lim[n->inf] (1 + 1/n)^n = e, but obviously we can write lim[n->inf] (1^n) = 1.
@shadowatom
@shadowatom 6 жыл бұрын
I don't believe you need the limit in this case because 1 is a constant. Your first example is two functions with one approaching 1 and the other approaching infinity which results in an indeterminate form. Since 1 is a constant in this case and 1 to any power is 1, you can safely say 1^inf is still 1.
@gopalrajpurohit8016
@gopalrajpurohit8016 Жыл бұрын
Is it high school math?
@OonHan
@OonHan 6 жыл бұрын
#t-shirt
@Cloud88Skywalker
@Cloud88Skywalker 6 жыл бұрын
8:48, isn't 0.9999... = 1 ?
@tommero6584
@tommero6584 6 жыл бұрын
He only said a finite amount of 9s. Not infinitely many.
@sugarfrosted2005
@sugarfrosted2005 6 жыл бұрын
It's easy to make this rigorous just by taking limit supremum or limit of X^(T_n)
@animeshpradhan5683
@animeshpradhan5683 6 жыл бұрын
integrate1/ ((cosx)^5+(sinx)^5) please
@gabob6992
@gabob6992 6 жыл бұрын
Proof that : x^2-y^3=1
@loganallomes4305
@loganallomes4305 6 жыл бұрын
Do you know the numbers have an end? -Mohamed Ababou-
@pramodgupta2998
@pramodgupta2998 Жыл бұрын
10:55 *Dirty mind activated* If you know you know
@danielbenyair300
@danielbenyair300 6 жыл бұрын
1st! Infinity is NOT a bumber!!! therefore CANNOT be positive or negative!!! 2nd! 1 does NOT equle to 0.99999999999999999999999... !!!! 3rd! This 0.000000000001^10000000000000000000 also does NOT equle to 0 !!!!!! 4th! the slope near 1 is INFINITY!!!!!!
@Alex-dn7jq
@Alex-dn7jq 6 жыл бұрын
Wait, 1^inf=1? I thought that this was indefinite.
@Alex-dn7jq
@Alex-dn7jq 6 жыл бұрын
misotanni got it! Thx!
@romanhredil3799
@romanhredil3799 6 жыл бұрын
Actually, 1^inf is divergent.
@thomashaas4381
@thomashaas4381 6 жыл бұрын
"If 1 is variable" is a very strange way to put it. The most natural interpretation of the term 1^inf is lim {n->inf} (1^n) which indeed is 1. Any other interpretation of that term is bogus. In any case, infinity is no real number and therefore shouldn't be treated as such. For convenvience reasons however, it is usually interpreted as a limit in case there is only one infinity symbol used. Terms like inf/inf or (1+1/inf)^inf containing 2 or more infinities are ambiguous since there is no reason to assume that both infinities are obtained as the limit of the same function.
@danielbenyair300
@danielbenyair300 6 жыл бұрын
alex ubuntu (-1)^infinity is indefinite... not +1
@KnakuanaRka
@KnakuanaRka 6 жыл бұрын
That’s only in limits, where it represents the limit of a function in the form of one expression to the power of a second expression, where those expressions tend towards 1 and i at the limit, such as the (1+1/x)^x limit used to evaluate e. As the value x approaches i, the base of that formula approaches 1, but never truly hits it, so the value doesn’t get stuck at 1. In this case, there’s no limit, so the base is exactly 1, and all the exponentiation isn’t he World makes no difference.
@deenaaalkotb
@deenaaalkotb 6 жыл бұрын
did you figure my problem out
@blackpenredpen
@blackpenredpen 6 жыл бұрын
I have a vid on rotation a polar curve kzbin.info/www/bejne/Z4jcomOnmdWMp7s For the parametric version, Sen Zen has done an awesome video on that already. kzbin.info/www/bejne/eIHKomV3qsmie9E
@deenaaalkotb
@deenaaalkotb 6 жыл бұрын
thank you for that
@deenaaalkotb
@deenaaalkotb 6 жыл бұрын
but I have another problem if f(x)=f(y) then x=y but sometimes it doesn't work like sin(3π/2)=sin(-π/4) what happened
@l3th4lv1p3r
@l3th4lv1p3r 6 жыл бұрын
deena alkotb f(x) = f(y) implies x=y is only true if f is an injection function en.m.wikipedia.org/wiki/Injective_function
@deenaaalkotb
@deenaaalkotb 6 жыл бұрын
@@l3th4lv1p3r thank you it was really nice 😎
@boda1948
@boda1948 5 жыл бұрын
I would like to call this "the binary function" because it only has 2 y values 1 and 0
@gg.3812
@gg.3812 6 жыл бұрын
At -1 it s not defined, it does not diverge
@ffggddss
@ffggddss 6 жыл бұрын
Terminology to describe that behavior differs. In my university courses long ago, any failure to converge was said to diverge. Nowadays a distinction is sometimes made between escaping to ±∞ ("divergence"), and failure to converge that is bounded ("oscillation"). In both cases, the function is said to be undefined at such a point. Fred
@rudboy9599
@rudboy9599 6 жыл бұрын
Got my master's last year. It's still the same. If it doesn't converge, it is said to diverge.
@Ben.2000
@Ben.2000 6 жыл бұрын
inf broke math. #yay!
@romanhredil3799
@romanhredil3799 6 жыл бұрын
but I can say 1^inf=e, or e^2, or e^3, or any positive number so 1^inf also diverges i'll show you the proof: You know this formula: lim_n→inf ((1+1/n)^n)=e so, if you plug infinity, you'll get (1+1/inf)^inf, 1/inf=0, so we just have 1^inf, but by the formula higher, we get this: 1^inf=e, but 1^inf=(1^n)^inf=(1^inf)^n=e^n, where n is any real number, and we get 1=n, which destroys all maths so, you must agree that 1^inf is divergent
@iabervon
@iabervon 6 жыл бұрын
Roman Hredil' When he says "a solid 1", he means that the base is a constant. That is, it's flatter than any other function. In particular, ln (base)
@adityatripathi1648
@adityatripathi1648 6 жыл бұрын
Someone's watching Doraemon in background
@PhasmidTutorials
@PhasmidTutorials 6 жыл бұрын
@weinsim3856
@weinsim3856 6 жыл бұрын
2nd
@EchoHeo
@EchoHeo 6 жыл бұрын
LOL i'm doing MATHS
@ogorangeduck
@ogorangeduck 6 жыл бұрын
2:41 hehe 太胖
@firstnamelastname7941
@firstnamelastname7941 6 жыл бұрын
I’m pretty sure 1^infinity is an indeterminate form.
@thomashaas4381
@thomashaas4381 6 жыл бұрын
There is no reason why it should be. No matter how you resolve the infinity as a limit of some function, you will end up with lim {n -> inf} 1^g(n) = 1 (for any g(n) which diverges towards infinity as n goes towards infinity). In the video you have g(n) = n, in which case you can directly see that the limit is 1 since it is the limit of the constant sequence 1^1, 1^2, 1^3, ...
@redsalmon9966
@redsalmon9966 6 жыл бұрын
Anything outside of this picture, the function just, NOT GOOD ANYMORE
@羅月鳳-v3t
@羅月鳳-v3t 6 жыл бұрын
gj4vm,6xl3g 數學老師?
@OonHan
@OonHan 6 жыл бұрын
Sell t-shirts!!!
@Bhamilton-ws4go
@Bhamilton-ws4go 6 жыл бұрын
#YAY
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