Post a New Question

Calculus

posted by on .

lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)
(x,y)->(1,0)


alright i approached this at y=0. and got the limit as 3. where else can i approach this from to see if the limit really exist which im hoping it doesn't. i tried y=x-1 and that was a mess and i also tried y=mx which did nothing.

  • Calculus - ,

    Hmmm. . . Looks like a removable singularity to me.

    x^3 - 1 = (x-1)(x^2 + xy + y^2)

    x^3+2x^2y+xy^2-x^2-2xy-y^2
    = x(x+y)^2 - (x+y)^2
    = (x-1)(x+y)^2

    so, f(x,y) = (x^2+xy+y^2)/(x+y)^2 for x≠1

    limit as (x,y) -> (1,0) = (1)/(1)^2 = 1

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question