习题7-11.判定下列平面点集中哪些是开集、闭集、区域、有界集、无界集？并指出集合的边界.（1）(x,y)x0,y0；（2）(x,y)1x2y24；（3）(x,y)yx2；（4）(x,y)x2(y1)21且x2(y2)24.解（1）集合是开集，无界集；边界为{(x,y)x0或y0}.（2）集合既非开集，又非闭集，是有界集；边界为{(x,y)x2y21}{(x,y)x2y24}.（3）集合是开集，区域，无界集；边界为{(x,y)yx2}.（4）集合是闭集，有界集；边界为{(x,y)x2(y1)21}{(x,y)x2(y2)24}2.已知函数f(u,v)uv，试求f(xy,xy).解f(xy,xy)xy(xy).3.设f(x,y)x4y42xy，证明：f(tx,ty)t2f(x,y).解f(tx,ty)tx4ty42t2xyt2x4y42t2xyt2x4y42xyt2f(x,y).yx2y24.设f(x0)，求f(x).xxy212yx2y2x解由于f，则fx1x2.xx15.求下列各函数的定义域：x2y2y（1）z；（2）zln(yx)arcsin；x2y2xx2y2（3）zln(xy)；（4）z1；a2b2z（5）zxy；（6）uarccos.x2y2解（1）定义域为(x,y)yx；（2）定义域为(x,y)xyx；（3）定义域为(x,y)xy0，即第一、三象限（不含坐标轴）；x2y2（4）定义域为(x,y)1；a2b2（5）定义域为(x,y)x0,y0,x2y；（6）定义域为(x,y,z)x2y2z20,x2y20.6.求下列各极限：x2xyy21cosx2y2（1）lim；（2）lim；(x,y)(2,0)xy(x,y)(0,0)ln(x2y21)1sin(xy)（3）lim(x2y2)sin；（4）lim；(x,y)(0,0)xy(x,y)(2,0)y1（5）lim(1xy)x；（6）lim(x2y2)exy.(x,y)(0,1)(x,y)(,)x2xyy24解：（1）limf(2,0)2；(x,y)(2,0)xy21u1cosx2y21cosu21（2）limlimlim；(x,y)(0,0)ln(x2y21)u0ln(1u)u0u211（3）因为lim(x2y2)0，且sin1有界，故lim(x2y2)sin0；(x,y)(0,0)xy(x,y)(0,0)xysin(xy)sin(xy)（4）limlimx212；(x,y)(2,0)y(x,y)(2,0)xy11y（5）lim(1xy)xlim(1xy)xye1e；(x,y)(0,1)(x,y)(0,1)(x2y2)(xy)2（6）当xN0，yN0时，有0，exyexyxy2u22u2而limlimlimlim0(x,y)(,)exyueuueuueu按夹逼定理得lim(x2y2)exy0.(x,y)(,)7.证明下列极限不存在：xy（1）lim；(x,y)(0,0)xyx2y,x2y20,（2）设f(x,y)x4y2limf(x,y).(x,y)(0,0)0,x2y20,证明（1）当(x,y)沿直线ykx趋于(0,0)时极限xyxkx1klimlim(x,y)(0,0)xyx0xkx1kykx与k有关，上述极限不存在.（2）当(x,y)沿直线yx和曲线yx2趋于(0,0)有x2yx2xxlimlimlim0，(x,y)(0,0)x4y2x0x4x2x0x21yxyxx2yx2x2x41limlimlim，(x,y)(0,0)x4y2x0x4x4x02x42yx2yx故函数f(x,y)在点(0,0)处二重极限不存在.8.指出下列函数在何处间断：1（1）zln(x2y2)；（2）z.y