5、1≤x+y≤4},2222边界为{(x,y)
6、x+y=1或x+y=4}.2(3){(x,y)
7、y>x};22解开集,区域,无界集,导集为{(x,y)
8、y≥x},边界为{(x,y)
9、y=x}.2222(4){(x,y)
10、x+(y−1)≥1}∩{(x,y)
11、x+(y−2
12、)≤4}.解闭集,有界集,导集与集合本身相同,2222边界为{(x,y)
13、x+(y−1)=1}∪{(x,y)
14、x+(y−2)=4}.22x2.已知函数f(x,y)=x+y−xytan,试求f(tx,ty).y22tx解f(tx,ty)=(tx)+(ty)−(tx⋅()ty)⋅(tan)ty=t2⎛x2+y2−xytanx⎞⎟=t2f(x,y).⎜⎝y⎠3.试证函数F(x,y)=lnx⋅lny满足关系式:F(xy,uv)=F(x,u)+F(x,v)+F(y,u)+F(y,v).证明F(xy,uv)=ln((x,y)⋅ln(uv)=(lnx+lny)(lnu+lnv)=lnx⋅lnu+
15、lnx⋅lnv+lny⋅lnu+lny⋅lnv=F(x,u)+F(x,v)+F(y,u)+F(y,v).wu+v4.已知函数f(u,v,w)=u+w,试求f(x+y,x−y,xy).xy(x+y)+(x−y)解f(x+y,x−y,xy)=(x+y)+(xy)xy2x=(x+y)+(xy).5.求下列各函数的定义域:2(1)z=ln(y−2x+1);2解要使函数有意义,必须y−2x+1>0,2故函数的定义域为D={(x,y)
16、y−2x+1>0}.11(2)z=+;x+yx−y解要使函数有意义,必须x+y>0,x−y>0,故函数的定义域为D={(x,y)
17、x+y>0,x−y>0}.(3
18、)z=x−y;2解要使函数有意义,必须y≥0,x−y≥0即x≥y,于是有x≥0且x≥y,2故函数定义域为D={(x,y)
19、x≥0,y≥0,x≥y}.x(4)z=ln(y−x)+;1−x2−y222解要使函数有意义,必须y−x>0,x≥0,1−x−y>0,22故函数的定义域为D={(x,y)
20、y−x>0,x≥0,x+y<1}.u=R2−x2−y2−z2+1(R>r>0);(5)x2+y2+z2−r222222222解要使函数有意义,必须R−x−y−z≥0且x+y+z−r>0,22222故函数的定义域为D={(x,y,z)
21、r22、z222解要使函数有意义,必须x+y≠0,且
23、
24、≤1即z≤x+y,x2+y222222故函数定义域为D={(x,y,z)
25、z≤x+y,x+y≠0}.6.求下列各极限:1−xy(1)lim;22(x,y)→)1,0(x+y1−xy1−0解lim==1.(x,y)→)1,0(x2+y20+1ln(x+ey)(2)lim;(x,y)→)0,1(x2+y2ln(x+ey)ln(1+e0)解lim==ln2.(x,y)→)0,1(x2+y212+022−xy+4(3)lim;(x,y)→)0,0(xy2−xy+42(−xy+4)(2+xy+)4解lim=lim(x,y)→)0,0(xy(x,
26、y)→)0,0(xy2(+xy+)4−11=lim=−.(x,y)→)0,0(2(+xy+)44xy(4)lim;(x,y)→)0,0(xy+1−1xyxy(xy+1+)1解lim=lim(x,y)→)0,0(xy+1−1(x,y)→)0,0((xy+1+1)(xy+1−)1xy(xy+1+)1=lim=limxy+1+)1=2.(x,y)→)0,0(xy(x,y)→)0,0(sin(xy)(5)lim;(x,y)→)0,2(ysin(xy)sinxy解lim=lim⋅x=1⋅2=2.(x,y)→)0,2(y(x,y)→)0,2(xy1−cos(x2+y2)(6)lim.22x2y
27、2(x,y)→)0,0((x+y)e1−cos(x2+y2)1−cos(x2+y2)1解lim=lim⋅lim22x2y222x2y2(x,y)→)0,0((x+y)e(x,y)→)0,0(x+y(x,y)→)0,0(e1−costsint=lim=lim=0.t→0tt→017.证明下列极限不存在:x+y(1)lim;(x,y)→)0,0(x−y证明如果动点p(x,y)沿y=0趋向(0,0),x+yx则lim=lim=1;(x,y)→)0,0(x−yx→0xy=0如
