I will do more calc 3 tutorials this year! Subscribe to support because my head is going to hurt 😆
@ranjithkumararunachalam38442 ай бұрын
Certainly ...😊
@thundercraft04962 ай бұрын
I was waiting for more calc 3 tutorial in your channel
@pollensalta2 ай бұрын
Please, do more vids on this limits. I always learn a lot thanks to you!!!!
@footballCartoon912 ай бұрын
May I ask a question, is it valid if someone plug in any number that is near to 0 rather than 0 itself into the formula when we talk about limit in mathematics. For example, if i plug in just a random number between 1 and 0 let say 0.5, 0.1 or 0.2 which one is valid and which one is not when we talk about limit in mathematics.
@cdkw22 ай бұрын
Take it easy upon yourself man, thanks for the great content
@guidichris2 ай бұрын
My head hurts. Hadn't thought about multivariable limits in 40 years.
@bitoty93572 ай бұрын
continue making these multivariable calc videos, they are great, and would be nice if you make videos about vector calc too
@sebas314152 ай бұрын
I love this! Why dont you do calc 3 on your main channel?
@bprpcalculusbasics2 ай бұрын
I want my main channels to be math for fun and my side channels to focus on tutorials and answering questions from the viewers.
@v2704982 ай бұрын
It'll be nice making a video about when polar coordinates limits work
@tobybartels84262 ай бұрын
In principle, you can use l'Hôpital's Rule in multivariable limits, although it usually doesn't get you anywhere. Instead of taking the _derivatives_ of the numerator and denominator, take their _differentials._ (Even in one-variable problems, this works, at the cost of an extra step; for example, to take the limit of (eˣ − 1)/(sin x) as x → 0, you check that l'H applies and change it to (eˣ dx)/(cos x dx), which simplifies to eˣ/(cos x), so the limit is 1.) In the first example from the video, l'H applies (in the form 0/0) and produces (6xy dx + 3x² dy)/(2x dx + 2y dy), but now we can't cancel anything, so it doesn't help. However, consider the limit of (x² + 2xy + y²)/(eˣ⁺ʸ − 1) as (x,y) → (0,0). Again, l'H applies (0/0), and now it produces (2x dx + 2y dx + 2x dy + 2y dy)/(eˣ⁺ʸ dx + eˣ⁺ʸ dy). But this time, we can factor and cancel dx + dy, simplifying to (2x + 2y)/(eˣ⁺ʸ), so the limit is 0.
@MathCuriousityАй бұрын
Somebody told me it is illegal to use lhopital for multivariate limits. Can you confirm they are wrong? If so why!?
@tobybartels8426Ай бұрын
@@MathCuriousity : Calling it ‘illegal’ suggests that there's something that you could do that would give a wrong answer. Maybe they're talking about doing it with a _partial_ derivative; I'll agree that that's illegal (it can make you think that the limit exists when it doesn't, or the other way around, although it will never tell you 3 when the limit is really 2, for example). But if you do it with differentials, then it's usually just _useless;_ if the differentials don't cancel, then it just doesn't give you an answer at all.
@MathCuriousityАй бұрын
@@tobybartels8426 thanks so much!
@tobybartels8426Ай бұрын
@@MathCuriousity : You're welcome!
@sreser1112 ай бұрын
Now that was fun to watch. Thank you much.
@carultch2 ай бұрын
He has it in Chinese as well. That was also fun to watch.
@bprpcalculusbasics2 ай бұрын
Thank you!
@LiamFranklinFarang2 ай бұрын
This is great. Please post more like it
@bprpcalculusbasics2 ай бұрын
Thanks. Will do!
@gregstunts3472 ай бұрын
This is similar to the case for x^y as (x, y) approaches (0, 0). Using polar coordinates, the limit approaches 1, or 0 when the angle is equal to pi/2 or 3pi/2. But if you use the substitution x = t, and y = c/ln(t) as t approaches 0, the limit can approach any positive value.
@gregstunts3472 ай бұрын
Of course though, it’s different to the cases in the video, because we are asking what potential values the limit can take. Rather than the actual value of the limit (as it DNE).
@MathCuriousityАй бұрын
Could we have stopped at 6:40 when we realized depending on theta we might get 0/0 ?
@gregstunts347Ай бұрын
@@MathCuriousity No. It just means that using polar coordinates isn’t enough to determine whether the limit exists.
@yplayergames79342 ай бұрын
I learned this in my Calculus 2 classes, and this was the most fun topic to learn
@bodesshorts86402 ай бұрын
Keep going on these calc 3 videos it will help me a lot 👍🏻
@DeJay72 ай бұрын
Sorry, but if I graph the second function, as (3x^2*y)/(x^4+y^2) = 0, (x,y) does approach (0,0). And that's what the first function also did. Maybe I don't know something (because I haven't done multivariable limits ever), but this seems weird.
@alessiodaini79072 ай бұрын
lim(0,0) f(x,y) ≠ lim x->0,y->0 f(x,y) ≠ lim x->0 f(x). but by a substitution as y=x that's y=0 the limit can be used to verify is correct or not, because if it exists, it should get the same limit by using other geometrical flat surfaces, indeed it isn't verified for y=x², for example. Indeed there isn't a solution and his calculations are correct. So there isn't a solution of that limit, because there's a theorem that says the limit is necessarily unique.
@alessiodaini79072 ай бұрын
ah, I studied this, here it's called analysis 2 and they were the first lessons. It's easier to prove a limit doesn't exist than verifying that it's calculable. Anyway doing this is like verifying if the limit at the left and the right of the point is equal or not, when you use a variable. It isn't enough to prove that's correct, but it's enough to verify that doesn't exist the limit in case they're different
@juanavelinobobadillabravo2 ай бұрын
Sr: Aprovecho para expresarle una inquietud, es que hace poco tocó el tema de los momentos de inercia y la suma de lo que ud demostró es el momento polar de inercia de un Área. Pero se debe calcular el momento de inercia (o el segundo momento del Área) respecto a un eje conocido y he aquí que debe seguir con el Teorema de Steiner, donde el Centroide del Área se encuentra respecto a un eje paralelo. Esto es de imperdonable importancia pues a partir de esto se determina:bh^3/12 (Ej del rectángulo respecto a su C de gravedad, etc, etc)y otros que espero ud los mencioné.
@raznologija7452 ай бұрын
Excellent explanation
@nicholarucitadhamma60002 ай бұрын
I'm confused on why Lim ((3x²•0)/(x⁴+0²)) = 0 x→0 Why is it zero and not some other value like infinity or just straight up indeterminate?
@bprpcalculusbasics2 ай бұрын
We do it inside out. 0/x^4 simplifies to 0. Then we look at the limit as x goes to 0. But for the 0/sin^2(theta) case is trickier. Theta can be anything such as 0, pi, 2pi or so. Those values will make the bottom 0. But we can’t make any conclusion in this case.
@nicholarucitadhamma60002 ай бұрын
@@bprpcalculusbasics oh right, that's a rule... Thank you for the explanation!
@davidcroft952 ай бұрын
Is there an equivalent theorem in multivariable analysis for L'Hopital rule?
@abacaabaca81312 ай бұрын
I think to learn math, someone really needs to buy a white board and a board marker that can be erased so that in the long term they will not waste money buying a pile of papers or a pile of books.
@MathCuriousityАй бұрын
Would we be able to use lhopitals at 6:30 or is lhopitals never allowed for multivariate limits?
@allmight8012 ай бұрын
Do you know how to change order of integration when in polar coordinates, dr dtheta and dtheta dr? What about cylindrical and spherical coordinates as well?
@BilalAhmed-on4kd2 ай бұрын
why can we use different "paths" here
@F_A_F1232 ай бұрын
by definition of the limit
@BilalAhmed-on4kd2 ай бұрын
@@F_A_F123 how is it related
@F_A_F1232 ай бұрын
@@BilalAhmed-on4kd ???? we take those paths to prove that the limit doesn't exist, wdym how is it related to the definition of the limit?
@BilalAhmed-on4kd2 ай бұрын
@@F_A_F123 i mean how is the fact that multiple paths give different results related to the non existence of the limit, why are we even allowed to take multiple paths
@F_A_F1232 ай бұрын
@@BilalAhmed-on4kd if you get closer to (0, 0) , a function that has a limit at that point will approach a certain value. But you can get closer to (0, 0) in different ways. If you do it one way, do it another way and the function approaches a certain value in both cases, but those values are different - a function doesn't have a limit at that point. This works in 1 dimension too: f(x) = {1, when x > 0; 0 when x < 0 Then: lim(x→0) f(x): if we take the path x > 0, f(x) approaches 1; if we take the path x < 0, f(x) approaches 0. That means f(x) doesn't have a limit at 0
@MathCuriousityАй бұрын
Can you confirm that it is illegal to use L’hopital’s rule for multi variable limits? I got one person saying you can and another that you can’t! Why can or can’t we? Thanks @bprpcalculusbasics