Bro you legit need to make more multivariable calc videos I have never seen a better educator You, sal khan and 3blue1brown are legit gems when it comes to giving the intuition behind the math Thanks a lot man
Teachers like you present math in color, and I don't say that just because this is a video in in color. There is energy in the words you say that illuminates the abstract concepts. Thank you.
5:53 I would point out that the numerator factors by grouping. I think that would clear some of the ambiguity of guessing and checking how the multivariable polynomial could be factored. In general, factoring a multivariable polynomial involves some kind of factoring by grouping, a concept familiar to many from algebra class.
Thanks and love u ,no one taught better than u I am far away from my class syllabus because to take classes from u seems to be watching movie in theatre
6:30 here's a explaination why its allowed to cancel these out without disturbing the limit (this applies for single variable limits as well): lim (x,y)->(3,2) (y^2+2)(x-3)/y(3-x) = (y^2+2)/y * -(x-3)/(x-3) the first term is continuous at 3,2 so the limit is just value of function, equals 11/3. The second term is pretty much a piecewise function of form : {undefined for x=3, 1 for any other x}. Now, recall the scary epsilon delta definition of limit, if you can find arbitrarily small region around x0 (not necessarily including x0), where value of function is arbitrarily close to some value L, then the L is a limit at x0. From the piecewise form we clearly see the function is 1 everywhere but at x=3, so clearly the limit will be L=1. So by cancelling we are removing a discontinuity, I actually don't exactly know why algebraic cancelling works, maybe it just happens to, but it can be justified by this longer proof.
You have great videos but there is a mistake here. You have (x,y) going to (2,3) meaning x is 2 and y is 3. Substituting those values into the denominator 3(3) - 2(3) = 3, not zero. Likewise, the numerator goes to -11. I think that perhaps you meant (x,y) is going to (3,2)?
Hi, I had a question, but firstly great video and thank you for making series like these :) So I was wondering in the conjugate example, why do we need that restriction, should we be able to still evaluate the limit in the same way without it, and if not, how do we know when we don't have restrictions, that the limit is undefined? Also another question I had was, if a limit goes to multiple values depending on the path, is there a way we can define the set of all values it approaches? It seems that it is against the spirit of limits to just say it doesn't exist and then job finished, isnt it is better to make a distinction between a limit that really doesn't exist anywhere on the function and a limit that exists for a set of restrictions and can be defined in a set of real numbers?
At 6:13, in order to explain the cancellation would it be more straightforward to factorise a -1 from the (x-3) or (3-x) to cancel the the notation out? It seems more intuitive to explain that way. A huge fan of the series by the way. Thanks for putting together!
i did the second example in factoring the numerator with (x + y - 4) = (sqrt(x + y) - 2)(sqrt(x + y) + 2) and then canceling the (sqrt(x + y) - 2) in the bottom it turns out to be the same thing just like to share thanks for the video
What could we do in situations when we cannot apply these tricks? For example if we have some ugly function/limit like (y*x^2)/(x^4 +y^2) as (x,y) tends to (0,0)? And thanks a lot for your channel and your work, it's wonderful
Great video sir.🎉🎉 Although I thought you were going to teach how to prove that a limit exists when we have tried all those paths and we got the same answers. Do you have a video for that?
I have a question... Why can we calculate the first limit simply by plugging in the values of x and y? Shouldn't we check if the limit exists by restricting the function?
Nice vid! Would you consider making more videos on multivariable limits or are the 2 in your calc 3 playlist the only ones you plan to have? Either way thanks!
Only got a couple right now as I work on vector calc and ODEs. However I know that calc 3 is still quite light in a few places so I do mean to come back and fill in as time goes on (it was originally for a flipped classroom, not an entire online course)
5:14 I think the numerator and the denominator are not going to zero at that point!🙄
i think that the mistake is in (x,y)->(2,3) , it would be (x,y)->(3,2) for the indetermitation.
good videos! thanks :)
(sorry by my english)
@@martivij He noticed in the description
@@huseyn.asadullah 😅 oops
Thank you, this was driving me crazy
Thanks, I was feeling so gaslit. :D I agree he probably mixed up the 2 and the 3.
Bro you legit need to make more multivariable calc videos
I have never seen a better educator
You, sal khan and 3blue1brown are legit gems when it comes to giving the intuition behind the math
Thanks a lot man
You must try professor leonard lectures then
blackpenredpen!! makes good vids too
@@coleyoutubechannel He usually focuses on single variable though I don't know if he even has any content on MVC
Teachers like you present math in color, and I don't say that just because this is a video in in color. There is energy in the words you say that illuminates the abstract concepts. Thank you.
In your 2nd example, isn’t the bottom 3 (3*3-2*3)?
It's awesome that you are making such helpful math videos through this pandemic - thank you!!
5:53 I would point out that the numerator factors by grouping. I think that would clear some of the ambiguity of guessing and checking how the multivariable polynomial could be factored. In general, factoring a multivariable polynomial involves some kind of factoring by grouping, a concept familiar to many from algebra class.
This series is really helpful! Appriciate it a ton!!
Thanks and love u ,no one taught better than u I am far away from my class syllabus because to take classes from u seems to be watching movie in theatre
6:30 here's a explaination why its allowed to cancel these out without disturbing the limit (this applies for single variable limits as well):
lim (x,y)->(3,2) (y^2+2)(x-3)/y(3-x) = (y^2+2)/y * -(x-3)/(x-3)
the first term is continuous at 3,2 so the limit is just value of function, equals 11/3.
The second term is pretty much a piecewise function of form :
{undefined for x=3, 1 for any other x}.
Now, recall the scary epsilon delta definition of limit, if you can find arbitrarily small region around x0 (not necessarily including x0), where value of function is arbitrarily close to some value L, then the L is a limit at x0. From the piecewise form we clearly see the function is 1 everywhere but at x=3, so clearly the limit will be L=1.
So by cancelling we are removing a discontinuity, I actually don't exactly know why algebraic cancelling works, maybe it just happens to, but it can be justified by this longer proof.
Really helpful!! Please keep uploading such informative videos..
Sir there was three in the denominator. So we could plug the values of x and y directly, i guess. (3y-xy)
You have great videos but there is a mistake here. You have (x,y) going to (2,3) meaning x is 2 and y is 3. Substituting those values into the denominator 3(3) - 2(3) = 3, not zero. Likewise, the numerator goes to -11. I think that perhaps you meant (x,y) is going to (3,2)?
TREFOR, YOU ARE A LEGEND!!!
Hi, I had a question, but firstly great video and thank you for making series like these :)
So I was wondering in the conjugate example, why do we need that restriction, should we be able to still evaluate the limit in the same way without it, and if not, how do we know when we don't have restrictions, that the limit is undefined? Also another question I had was, if a limit goes to multiple values depending on the path, is there a way we can define the set of all values it approaches? It seems that it is against the spirit of limits to just say it doesn't exist and then job finished, isnt it is better to make a distinction between a limit that really doesn't exist anywhere on the function and a limit that exists for a set of restrictions and can be defined in a set of real numbers?
I think you have to decline it at first but can allow that x + y = 4, when you canceled out the one term.
Great critical thinking there! May you keep it up and hope Trefor will reply soon.
@@nonentity168 thanks man, but it's 2 years later lol, I don't think he's replying
Thank you so much sir for this wonderful course 🔥
"We don't care what's happening at the point, we care what is happening around the point."
!!!!!!!
You missed two more ways. First, by using the squeeze theorem and second by using the epsilon-delta definition :D.
At 6:13, in order to explain the cancellation would it be more straightforward to factorise a -1 from the (x-3) or (3-x) to cancel the the notation out? It seems more intuitive to explain that way.
A huge fan of the series by the way. Thanks for putting together!
That works too, and thank you!
i did the second example in factoring the numerator with (x + y - 4) = (sqrt(x + y) - 2)(sqrt(x + y) + 2)
and then canceling the (sqrt(x + y) - 2) in the bottom
it turns out to be the same thing
just like to share
thanks for the video
What could we do in situations when we cannot apply these tricks? For example if we have some ugly function/limit like (y*x^2)/(x^4 +y^2) as (x,y) tends to (0,0)?
And thanks a lot for your channel and your work, it's wonderful
If you can’t manipulate it algebraically to use some more elementary limit laws, then it becomes pretty hard!
set y = x and see the limit isn't equal
Great video sir.🎉🎉
Although I thought you were going to teach how to prove that a limit exists when we have tried all those paths and we got the same answers.
Do you have a video for that?
Example 2 limit is (x,y)-->(3,3)
I have a question... Why can we calculate the first limit simply by plugging in the values of x and y? Shouldn't we check if the limit exists by restricting the function?
that's fine UNLESS doing so means something like division by zero
I'm new to multivariable calc. At 5:14, could you use L'Hopital's Rule?
Late response? But no, you can't as LH's only applies to single-variable functions
How can we solve? (X^2 y)/(x^2 + 3Y^2)
thanks a lot ♥
You're welcome 😊
Nice vid! Would you consider making more videos on multivariable limits or are the 2 in your calc 3 playlist the only ones you plan to have? Either way thanks!
Only got a couple right now as I work on vector calc and ODEs. However I know that calc 3 is still quite light in a few places so I do mean to come back and fill in as time goes on (it was originally for a flipped classroom, not an entire online course)
Thanks
Nice sir
Sir, can you make a video on the ε-δ approach?
Is there some extension of L'Hopital rule in higher variable calculus?
9:55 you assumed x+y was not 4 and then you pluged in so x+y=4?
why did you, in 3d example, said x + y =/ 4, but then in the last step you have that x + y = 4 ?
thaankkk you
Tnx sir
🌷
Can I use two different path to show that limit do exist
sadly not!
hey insufficient likes for such video!!!!!!!