$\begingroup$ $(xy)^2 > 0 \implies x^2 y^2 2xy > 0\implies \frac 12 (x^2y^2) > xy$ $\endgroup$ – Doug M Apr '17 at 06 $\begingroup$ Fantastic!Limit as (x,y) approaching (0,0) of (xy)/ (x^2y^2) \square!Where f(0;0) = 0 and f(x;y) for (x;y) 6= (0 ;0) is given by i) j x j j y j i) pxy x2y2 ii) xy x2y2 iii) x4¡y2 x4y2 iv) x2y x4y2 2
Tan 1 X 2 Y 2 X 2 Y 2 A Novocom Top
