f(x,y)=⎩⎨⎧x+x2+y2xy0(x,y)=(0,0)(x,y)=(0,0)
fx=1+(x2+y2)2y(x2+y2)−xy⋅2x=1+(x2+y2)2−x2y+y3
lim(x,y)→(0,0)fx=1
fy=(x2+y2)2x(x2+y2)−xy⋅2y=(x2+y2)2x3−xy2
lim(x,y)→(0,0)fy=0
よって原点で連続でない
f(x,y)=⎩⎨⎧x+x2+y2xy0(x,y)=(0,0)(x,y)=(0,0)
fx=1+(x2+y2)2y(x2+y2)−xy⋅2x=1+(x2+y2)2−x2y+y3
lim(x,y)→(0,0)fx=1
fy=(x2+y2)2x(x2+y2)−xy⋅2y=(x2+y2)2x3−xy2
lim(x,y)→(0,0)fy=0
よって原点で連続でない
解説: r31bn1z