I don't know why did you unnecessarily extended the problem, after substituting x, y and z you could see x²+y²+z²=p² which you noted on left side of board. But still instead of replacing the denominator with p² you decided to solve the long trig expression. Which killed the entire purpose if teaching "how" the substitution came in mind, before solving the problem.
I started learning this stuff literally 50 years ago but I still learn new things. This was a great problem and a very good solution. Your explanations are always very clear and easy to understand.
I like how they don’t assume prior knowledge at all. Some students find that annoying because they know certain things already, but they aren’t all students.
I found your videos recently and honestly they are some of the best math videos on TH-cam. You explain everything so clearly and have taught me so much ❤
Very interesting. If I were faced with how to approach this with all variables decreasing equally I would let X = Y = Z so look at what x^6 / (3X^2) does - which is basically X^4/3 as X goes to zero so zero. This trival simplification I would think shows if you shrink the sphere equally - you get convergence to zero. Very neat and accurate how you did it though - and a better general solution that my point reduction / simplification of the general question!
With spherical coordinates you immediately get that the bottom is just rho² like in your calculation it shows. Rho is the distance between the origin and (x, y, z) so, rho² = x² + y² + z²
One thing that I don't think people appreciate enough about multivariable limit is the sheer jump in burden of proof. When doing single variable limits, you only have 1 axis for your approach which simplifies things greatly. As soon as you get limits in 2 dimensions, the number of directions of approach become infinite (and that's only considering linear paths of approach -- there are of course more complicated paths). I think this is where the power of epsilon-delta shines.
It appears, intuitively, that the limit will be zero: choose x,y,z => epsilon, some very small number close to zero, denoted p. Then we have p^6 in the numerator and 3p^2 in the denominator which equals p^4 / 3. As p gets closer to zero, the ratio gets closer and closer to zero. I'm anxious to see your presentation on a more formal approach.
Is this allowed: Let y = ax, z = bx So the limit becomes ((a^2*b^2)x^6)/((a^2+b^2+1)x^3)= x^3 *(a^2 b^2)/(a^2 + b^2 + 1) as x goes to 0 which is just 0^3 times some positive constant = 0
No. You prove that on any straight line containing point (0,0,0) the limit is 0. But we can approach (0,0,0) on different curves, you should consider them all (infinitely many curves).
needlessly complicated: at 6:16, we have the equation rho = sqrt(x^2 + y^2 + z^2) at 6:32, denominator is x^2 + y^2 + z^2, which is equal to rho^2, do not need to write complicated expression and then show that this reduces to rho^2
Thank you Sir for this very interesting video. However, I have a question about ascribing x and y with new functions. As you showed, it is clear that x and y have to be dependent together after you did this. Would you explain your decision please Sir?
The simplification of the numerator is realy satisfying, but considering that cos²(theta)sin²(theta)cos²(phi)sin(phi)^4/ (cos²(theta)sin²(theta)+sin²(theta)sin²(theta)+cos²(phi)) Doesnt depend on rho could we just call it A(theta, phi) and rewrite the limit as lim rho->0 rho^4 * A(theta,phi) = A(theta, phi) lim rho-> 0 rho^4 = 0 ?
Knew it was most likely zero. I did a total of 5 cases in my head within 6 seconds and they all yielded 0. x = 0, y = 0, z = 0, ( x = y or y=x and x or y= 0). x = y = z This shows the limit tends to 0
Solution: Given that all variables go to 0, we can replace them with a single variable like "u", and get: lim u → 0 for (u²u²u²) / (u² + u² + u²) = lim u → 0 for u⁶/(3u²) = lim u → 0 for u⁴/3 => 0⁴/3 = 0/3 = 0
@@m.h.6470 If you're going to compare your calculation to the video: In the first 2 minutes, he explains your method as well as what your mistake is (he describes that as the method not taking every direction pointing at (0,0,0) into account), then he explains how his method of taking a sphere of all equidistant points does take every direction into account, meaning that he debunked both of your comments within the first 5 minutes of the video that you're commenting under. That being said, one thing that he neglected to mention is that the function f_ρ(φ,θ)=ρ⁴cos²(θ)sin²(θ)sin⁴(φ)cos²(φ) needs to *uniformly* converge to 0 as ρ->0, but he did take that into account by mentioning that the term that is dependent on φ and θ isn't "infinite", meaning that it's a bounded function.
rho rho rho your ball, gently down the sphere
While I know I'll never use this, the limit solution was educational. I'm still living.
I don't know why did you unnecessarily extended the problem, after substituting x, y and z you could see x²+y²+z²=p² which you noted on left side of board. But still instead of replacing the denominator with p² you decided to solve the long trig expression. Which killed the entire purpose if teaching "how" the substitution came in mind, before solving the problem.
You're right, my mind didn't go there. It was a major blunder. It usually doesn't work that nicely. I guess that's why I didn't see it.
@@PrimeNewtons No problem try to solve JEE ADVANCED problems too they are just awesome.
I started learning this stuff literally 50 years ago but I still learn new things. This was a great problem and a very good solution. Your explanations are always very clear and easy to understand.
I like how they don’t assume prior knowledge at all. Some students find that annoying because they know certain things already, but they aren’t all students.
I am a bigg fan of you sir❤.
A student from India 😅
I found your videos recently and honestly they are some of the best math videos on TH-cam. You explain everything so clearly and have taught me so much ❤
I just love you Prime!
this is absolutely awesome, and it looks fun!!!
Very interesting. If I were faced with how to approach this with all variables decreasing equally I would let X = Y = Z so look at what x^6 / (3X^2) does - which is basically X^4/3 as X goes to zero so zero. This trival simplification I would think shows if you shrink the sphere equally - you get convergence to zero.
Very neat and accurate how you did it though - and a better general solution that my point reduction / simplification of the general question!
With spherical coordinates you immediately get that the bottom is just rho² like in your calculation it shows. Rho is the distance between the origin and (x, y, z) so, rho² = x² + y² + z²
One thing that I don't think people appreciate enough about multivariable limit is the sheer jump in burden of proof. When doing single variable limits, you only have 1 axis for your approach which simplifies things greatly. As soon as you get limits in 2 dimensions, the number of directions of approach become infinite (and that's only considering linear paths of approach -- there are of course more complicated paths). I think this is where the power of epsilon-delta shines.
It appears, intuitively, that the limit will be zero: choose x,y,z => epsilon, some very small number close to zero, denoted p.
Then we have p^6 in the numerator and 3p^2 in the denominator which equals p^4 / 3. As p gets closer to zero, the ratio gets closer and closer to zero.
I'm anxious to see your presentation on a more formal approach.
Great vidéo ❤
Is this allowed:
Let y = ax, z = bx
So the limit becomes ((a^2*b^2)x^6)/((a^2+b^2+1)x^3)= x^3 *(a^2 b^2)/(a^2 + b^2 + 1) as x goes to 0 which is just 0^3 times some positive constant = 0
No.
You prove that on any straight line containing point (0,0,0) the limit is 0. But we can approach (0,0,0) on different curves, you should consider them all (infinitely many curves).
needlessly complicated: at 6:16, we have the equation rho = sqrt(x^2 + y^2 + z^2)
at 6:32, denominator is x^2 + y^2 + z^2, which is equal to rho^2, do not need to write complicated expression and then show that this reduces to rho^2
very cool, typically I only ever used polar sub
Thank you Sir for this very interesting video. However, I have a question about ascribing x and y with new functions. As you showed, it is clear that x and y have to be dependent together after you did this. Would you explain your decision please Sir?
Wait, how would epsilon-delta work with multiple variables? If it's okay with you, could you make a video about that?
Amazing video💪 Id like to see more multivariable functiom stuff(limits/multiple integrals/...). Thanl god I have found your channel last year!
your voice is very calm
What is limit if numerator was (xy)?
How will you solve it?
Came for the math. Stayed for the "Chaa!" at the end 10:58
Very good. Thanks 🙏
L art pour l art
The denominator is rho*rho by definition
Rest is trivial
The simplification of the numerator is realy satisfying, but considering that cos²(theta)sin²(theta)cos²(phi)sin(phi)^4/
(cos²(theta)sin²(theta)+sin²(theta)sin²(theta)+cos²(phi))
Doesnt depend on rho
could we just call it A(theta, phi)
and rewrite the limit as lim rho->0 rho^4 * A(theta,phi) = A(theta, phi) lim rho-> 0 rho^4
= 0
?
im in grade 10 and multivariable calculas is beatifull when seen in play, where could i get more resources
If you power all the equation by minus one you can see it is equal to infinity very easily
6:02 I was skipping and saw this face that scared me 😂
I just assumed it was 0 because the numerator approaches 0 faster than the denominator
Somehow, I feel better now 😀
Numerator goes to zero like power of 6, denominator goes to zero like power 2, quotient goes to zero like power of 4 => limit 0.
:o wow
That's a lot of rho
لوبيتال
Knew it was most likely zero. I did a total of 5 cases in my head within 6 seconds and they all yielded 0. x = 0, y = 0, z = 0, ( x = y or y=x and x or y= 0). x = y = z
This shows the limit tends to 0
Solution:
Given that all variables go to 0, we can replace them with a single variable like "u", and get:
lim u → 0 for (u²u²u²) / (u² + u² + u²)
= lim u → 0 for u⁶/(3u²)
= lim u → 0 for u⁴/3
=> 0⁴/3 = 0/3 = 0
The problem with this is that it assumes that the limit exists. If it doesn't exist, then not all paths to (0,0,0) converge or yield the same result
@@kappasphere the same thing is true for the calculation in the video...
@@m.h.6470 If you're going to compare your calculation to the video:
In the first 2 minutes, he explains your method as well as what your mistake is (he describes that as the method not taking every direction pointing at (0,0,0) into account), then he explains how his method of taking a sphere of all equidistant points does take every direction into account, meaning that he debunked both of your comments within the first 5 minutes of the video that you're commenting under.
That being said, one thing that he neglected to mention is that the function f_ρ(φ,θ)=ρ⁴cos²(θ)sin²(θ)sin⁴(φ)cos²(φ) needs to *uniformly* converge to 0 as ρ->0, but he did take that into account by mentioning that the term that is dependent on φ and θ isn't "infinite", meaning that it's a bounded function.
I took 5 seconds to conclude it is 0. No need to switch x, y, z to sin and cos form.