Limit of multivariable function

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In this video, I showed how to evaluate a multivariable limit using spherical coordinates.
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rho rho rho your ball, gently down the sphere

luxxulyanite
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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.

herbertsusmann
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While I know I'll never use this, the limit solution was educational. I'm still living.

BartBuzz
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I found your videos recently and honestly they are some of the best math videos on YouTube. You explain everything so clearly and have taught me so much ❤

ThomasStupak
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I am a bigg fan of you sir❤.
A student from India 😅

RYedukrishnan-cnft
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this is absolutely awesome, and it looks fun!!!

ericfeng-ct
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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.

samrizzo
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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.

rachitchauhan
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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!

matthewkendall
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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

coreymonsta
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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.

nozack
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very cool, typically I only ever used polar sub

qwertyuiop
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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?

thuantran
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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

barryzeeberg
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Is this allowed:
Let y = ax, z = bx
So the limit becomes 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

ayushrudra
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The simplification of the numerator is realy satisfying, but considering that

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
?

glorrin
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Wait, how would epsilon-delta work with multiple variables? If it's okay with you, could you make a video about that?

nanamacapagal
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Amazing video💪 Id like to see more multivariable functiom stuff(limits/multiple integrals/...). Thanl god I have found your channel last year!

A.A
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Came for the math. Stayed for the "Chaa!" at the end 10:58

Dr.UldenWascht
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What is limit if numerator was (xy)?
How will you solve it?

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