求下列函数的极限：
(1)limx→0y→1(1-xy)/(x2-y2)；
(2)limx→0y→1/2arcsin√x+y
(3)limx→0y→0sin2(x2+y2)/(x2+y2)；
(4)limx→0y→0(2-√xy+4)/xy
(1)利用初等函数的极限得： limx→0y→1(1-xy)/(x2-y2)=(1-0•1)/(x2-y2)= (1-0•1)/(0-1)2=-1 (2)limx→0y→1/2arcsin√x+y=arcsin√(0+1/2)=arcsin(√2/2)=π/4 (3)limx→0y→0sin2(x2+y2)/(x2+y2)= limx→0y→0sin2(x2+y2)/[2(x2+y2)]•2 令x2+y2=u，则当x→0，y→0时，u→0，从而 limx→0y→0sin2(x2+y2)/(x2+y2)= =limu→0(sinu/u)•2=2 (4)令xy=u，则： limx→0y→0(2-√xy+4)/xy=limu→0(2-√u+4)/u= limu→0-u/(2+√u+4)u=limu→0-1/(2+√u+4)=-(1/4) ∴ limx→0y→0(2-√xy+4)/xy=-(1/4)
