1. The problem statement, all variables and given/known data Use the epsilon delta definition to show that lim(x,y) -> (0,0) (x*y^3)/(x^2 + 2y^2) = 0 2. Relevant equations sqrt(x^2) = |x| <= sqrt(x^2+y^2) ==> |x|/sqrt(x^2+y^2) <= 1 ==> |x|/(x^2+2y^2)? 3. The attempt at a solution This limit is true IFF for all values of epsilon > 0, there exists a delta such that: 0<sqrt(x^2+y^2)<delta ==> |(x*y^3)/(x^2+2y^2)| < epsilon When checking my answers, I'm stuck at this step: Can: |x|/sqrt(x^2+2y^2) <= 1 imply that |x|/(x^2+2y^2) will also be <= 1? I'm skeptical as it seems wrong, but if this step holds, then the rest of my proof can be as follows: (|x|/(x^2+2y^2))*|y^3| <= |y^3| = |(y^2)^(3/2)| <= (x^2 + y^2 ) ^ (3/2) < (delta)^3 thus let delta = (epsilon)^(1/3) and we get: sqrt(x^2+y^2)<epsilon^(1/3) ==> (x^2+y^2)^(3/2) < epsilon ==> (y^2)^(3/2) = y^3 < (x^2+y^2)^(3/2) < epsilon; Since |x|/(x^2+2y^2) <= 1, (|x|/(x^2+2y^2)) * |y^3| < |y^3| < epsilon; and we have f(x) < epsilon.