Question

Find the limit, if it exists, or type DNE if it does not exist. (Hint: use polar coordinates.) limit(x,y)→(0,0) (2x^3+2y^3) / (x^2+y^2)

Answer 1

x^3 + y^3 = (x + y) (x^2 - x y + y^2), which will not factor out any terms in common with (x^2+y^2)
You can try L'Hospital's rule, because it is of the indeterminant form 0/0: do you use the sum of partial derivatives with respect to x, and y for numerator and denominator?:

= limit 6x^2 + 6y^2 12x+12y
(x,y)->(0,0) ----------- = lim(...) --------- = 0
2x + 2y 2 + 2

Answer 2

0 is correct. but I have to have it in polar corrdinates

Answer 3

Let x = rcosa and y = rsin a,

limit(x,y)→(0,0) (2x^3+2y^3) / (x^2+y^2) =
limit(a,r)→(a,0) r^3(2cos^3(a)+2sin^3(a)) / (r^2) =
limit(a,r)→(a,0) r(2cos^3(a)+2sin^3(a)) = 0