In this video I use polar coordinates to calculate a limit in several variables. This is usually very hard to do, but polar coordinates simplify our work tremendously.
@0:57 I think you mean y=rsin(theta) on the whiteboard but you change it in the next line and when you are speaking so everything works out haha @3:57 when you multiple each term by r^2, you can make a tighter restriction on the lower bound because it was previously zero it's still zero so you don't even have to evaluate a limit for the lower bound.
The squeeze theorem is so important!!! I wish I understood this concept earlier, it somehow tangentially helped me understand L'Hospital's Rule, along with looking at the series expansions. Thank you for the upload!
Is the squeeze theorem necessary in this case though? Don't cos & sin have positive powers and therefore always larger than zero? I'm not sure - just observing.... Thank you Dr Peyam - love your videos!
Could you please consider on the limit of the function (y*x^3)/(y^2+x^6) at (0,0). I think polar coordinate approach does not work in this case. After polar coordinate switch, I calculate the limit as 0. On the other hand, as aproaching to (0,0) along the path y=x^3, the limit is 1/2. Is there anything wrong that I calculated? My reasoning is "polar coordinate is just representing the linear aproach to (0,0), because we are keeping theta as constant and r is approaching to 0. Then, if limit does not related to theta, this means all linear paths to (0,0) give same limit. But this approach can not represent curvilinear path approaches." At this point you may ask "why don't we use y=mx switch, instead of polar coordinates" y=mx represent all linear paths except x=0. By this point of view, I am not agree with your spiral approach explaining as well. What do you think about my reasoning?
@@drpeyam But in the video your statement at 4:08 implies that this method tests all paths, but it apparently doesnt. If this does not test a curved path such as the one that the commenter here gave, how can we test all of the paths?
Dr peyam in my class my professor had us evaluate it like a substitution Like we do y=MX and y=kx² If the both go to the same value, then the limit exists otherwise it doesn't So...can we do this same thing that way as well? And factor and cancel out any terms?
Hey Dr. Peyam, couldn't we use a similar argument to say that if theta is close to k*pi , rsin(theta) tends to zero and so y tending to zero is equivalent to theta tending to kpi? Shouldn't we divide the limits into the two cases; when r tends to zero and also the case theta tends to kpi (in the y case, the x case would be theta tending to kpi/2)?
Hi Dr Peyam! lim xy (x,y) -> (0,0) __________________ sqrt(x^2+y^2) Would this limit evaluate to zero using the polar coordinates? lim rcos(θ) * rsin(θ) r -> 0 __________________ sqrt(r^2) lim rcos(θ) * rsin(θ) r -> 0 __________________ r lim r*cos(θ)*sin(θ) r -> 0 0 * cos(θ)*sin(θ) No matter what θ is, this will always evaluate to zero without any issue. Hence, the limit is 0. Did I do this correctly? Thank you so much for any feedback :)
I believe it is also possible to let x(n)=x0+a/n and y(n)=y0+b/n for arbitrary a and b and then let n--> infinity. If the result depends on your choice of a and b then the function is not continuous in (x0,y0). This works because you can have arbitrary paths depending on how you choose a and b.
I don't think that works as you must consider all the possible ways x and y could go to zero not just straight lines. Peyam's proof works as it doesn't even matter whether theta varies on each path due to the squeeze theorem.
Consider the limit as x and y go to 0 of 2x^2y/(x^4+y^2) if you let y=mx then this expression becomes 2mx^3/(x^4+m^2x^2)>2mx^3/(m^2x^2)=2x/m which has limit of zero. However, if you let y=x^2 then this expression becomes 2x^4/(2x^4)=1. You would be alright if you said y=m(x)x and then showed the limit was zero like theta in Peyam's video has to be a function of x and y.
It’s not a supreme method to find if a function has a limit or not. You can use this to prove limit does not exist, yeah, but not to prove that limit exists or not. There might be other curve through which we may get a different limiting value.
Hi I tried to evaluate a limit using this. The function is x²y / (x^4 + y^2). And using this my answer came out to be zero but this limit doesn't exist. How do we know where to apply polar coordinates to evaluate limits and where not?
I found the center of c onic sections by following methodology: Step-1:let f(x,y)=0 be the equation of the conic section Step-2:then I calculated the partial derivative of f(x,y) with respect to x and y respectively. Step-3:Then,l equated those partial derivatives to zero and got 2 simultaneous equations Step-4:At last l solve those equations for (x,y) which was later found to be the coodinates of the conic section. I want to know the actual truth. So,l need your Noble help
It's not weird, rho as radius is a valid international notation. Phi as an angle is, however, not very common, as it can be misinterpreted as spherical coordinates. Also I like weird notations too, it widens your culture of mathematics.
I didn't said that it wasn't valid though Saying weird was just an expression, I like the notation a lot actually We are at the current moment talking about multiple integrals and for the Jacobian he used Lagrange's notation for derivatives trying to keep it minimalistic He's from Vietnam
I knew this method and i loved it at first sight but i always was in doubt if this was legal to do now i can confirm that it is. But whats bothering me is if this is true (i dont doubt at all its perfectly done) why the hell in all courses keep on doing deltha epsilon to proove something wich it is much more easy to do in polar coordinates. If this "proove" that the limit is zero why is so popular the tradtitional hardwork? That question i really cannot answear the fact that we cannot deny is that in all courses do for this kind of limit the epsilon theta and thats the reason why i always was affraid to do this limit in polar coordinates i always think that there is maybe something hide that u cannot do. The statistics use a 99% epsilon delta in all courses the question is whyyyyy???? If this is perfectly done i feel like the statistics tell me dont ever think to do it this way but i dont kjow why , it seems to me perfectly done
As both variables approach the same value, I used the limit as x gos to 0 and switched y to x. I also got 0, is that a coincidence? Can I always do that?
It looks great but I always wondered why on earth considering y=mx at a fixed m approaching x=0 doesn't seem to be enough to conclude anything but at a fixed theta approaching r=0 works ,it must be an extra fact behind because in both cases we fix one variable and then approach the remaining one to zero but in one case works and in the other doesn't. And i think ..the reason is because in y=mx we cannot conclude anything but in polar we can ...lol we already know that,... the question is why
i think its because when you test y=mx, you only check along all straight lines. Therefore, you can't make any claims for other paths. However, using polar coordinates sets no constraints on the nature of your path
Lets see if I can manage. The limit is (x4-y4+4y4)/(x2+y2) = (x2-y2)+4y4/(x2+y2), so we get 4*Lim(y4/(x2+y2)), and we can do the same backwards to find: (4x4-3x4+3y4)/(x2+y2), and we get 4*Lim(x4/(x2+y2)), to be expected. Maybe now we can go to 4*Lim(x2/(1+y2/x2)), and see that for all y2/x2 not approaching -1, that it approaches zero.
mmmm let's see if we take z=x.y^2/(x^2+y^4) at (0;0) and we approach with x=0 then it goes to 0 but if we take x=y^2 the limit goes to 1/2 so the limit does not exist. However if we plug x=r.cost and y=r.sin (t) we end up the whole thing going to zero!! So....the polar method fails here there must be some hidden conditions I must to check before using this I think it's not so easy to plug polar coordinates and make r goes to 0(which I confess I did the same for several years until I stumbled with this example today!!!)
Portsmouth is refering to the fact that if he applies -r transform (before taking the limit operator), it should have been 0 as the bottom inequality, and not -r^2
What exactly did you smoke before this video? It's a short one indeed, but at first you messed up by writing r=cos(theta), but fortunately, you noticed it Then you wrote "y = r cos(theta)" while the line "x = r cos(theta)" was 2 cm above. I hoped you would have figured out it was sine when you replaced at 1:19 At 3:27, bottom boundary should be zero, if you apply *r^2 to all sides of the inequality.