事实上 △z=AAx+B△y+0(P) lim△z=0, 0 limf(x+△x,y+△y) △x→>0 △y→>0 =limf(x,y)+△x] 0 =∫(x,y) 故函数z=f(x,y)在点(x,y)处连续 可微的条件事实上 z = Ax + By + o(), lim 0, 0  = → z  lim ( , ) 0 0 f x x y y y x +  +   →  → lim[ ( , ) ] 0 = f x y + z → = f (x, y) 故函数z = f (x, y)在点(x, y)处连续. 二、可微的条件