Multivariable Calculus | Finding a limit with polar coordinates.

แชร์
ฝัง
  • เผยแพร่เมื่อ 12 ก.ย. 2024
  • We use polar coordinates to find the limit of a multivariable function.
    www.michael-pen...
    www.randolphcol...

ความคิดเห็น • 5

  • @nailabenali7488
    @nailabenali7488 4 ปีที่แล้ว +13

    Hey! For the last example ( the one you told us to think about it) I was wondering if we suppose that it is actually possible to have cos(teta)+sin(teta)=0 if we multiply by r we gonna find x+y=0 but this can't happen since it's not in the domain of the function itself correct?

    • @MichaelPennMath
      @MichaelPennMath  4 ปีที่แล้ว +12

      That is right! All points (x,y) such that x+y=0 are not part of the domain of this function. As such, the proper (epsilon/delta) definition of the limit would not consider these in the first place.

  • @AlexandreRibeiroXRV7
    @AlexandreRibeiroXRV7 ปีที่แล้ว +2

    A bit late to comment on this video, but for the last function we actually don't have a limit.
    Consider a parametrization of (x, y) = (t, t^n - t), where n is a positive real number; when t goes to 0, so will both x and y.
    Yet, if you plug these values into (x^2 + y^2)/(x + y), you will find that this evaluates to 2t^(2 - n) + t^n -2t, which depending on the value of n will give a different value for the limit:
    * 0 < n < 2 gives a limit of 0
    * n = 2 gives a limit of 2
    * n > 3 creates a division by 0
    Which therefore means that we cannot assign a value to that limit.

  • @l-unnamed802
    @l-unnamed802 ปีที่แล้ว

    But I don't understand why r must be from positive side, why not from all side sides?

    • @raphaeldirollo3229
      @raphaeldirollo3229 ปีที่แล้ว +4

      It’s because r designates the radius in polar coordinates so it can’t be < 0