作业帮1.画出方程表示的曲面：z= -（√(x^2+y^2)）2.证明极限lim [(x+y)/(x-y)]不存在x→0,y→03.求函数极限lim[(x+y)sin(1/x^2+y^2)],lim[(xy)/（√（xy+1））-1]x→0,y→0 x→0,y→0
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![1.画出方程表示的曲面：z= -（√(x^2+y^2)）2.证明极限lim [(x+y)/(x-y)]不存在x→0,y→03.求函数极限lim[(x+y)sin(1/x^2+y^2)],lim[(xy)/（√（xy+1））-1]x→0,y→0 x→0,y→0](/uploads/image/z/13283264-56-4.jpg?t=1.%E7%94%BB%E5%87%BA%E6%96%B9%E7%A8%8B%E8%A1%A8%E7%A4%BA%E7%9A%84%E6%9B%B2%E9%9D%A2%EF%BC%9Az%3D+-%EF%BC%88%E2%88%9A%28x%5E2%2By%5E2%29%EF%BC%892.%E8%AF%81%E6%98%8E%E6%9E%81%E9%99%90lim+%5B%28x%2By%29%2F%28x-y%29%5D%E4%B8%8D%E5%AD%98%E5%9C%A8x%E2%86%920%2Cy%E2%86%9203.%E6%B1%82%E5%87%BD%E6%95%B0%E6%9E%81%E9%99%90lim%5B%28x%2By%29sin%281%2Fx%5E2%2By%5E2%29%5D%2Clim%5B%28xy%29%2F%EF%BC%88%E2%88%9A%EF%BC%88xy%2B1%EF%BC%89%EF%BC%89-1%5Dx%E2%86%920%2Cy%E2%86%920+x%E2%86%920%2Cy%E2%86%920)
1.画出方程表示的曲面：z= -（√(x^2+y^2)）2.证明极限lim [(x+y)/(x-y)]不存在x→0,y→03.求函数极限lim[(x+y)sin(1/x^2+y^2)],lim[(xy)/（√（xy+1））-1]x→0,y→0 x→0,y→0
1.画出方程表示的曲面：z= -（√(x^2+y^2)）
2.证明极限lim [(x+y)/(x-y)]不存在
x→0,y→0
3.求函数极限lim[(x+y)sin(1/x^2+y^2)],lim[(xy)/（√（xy+1））-1]
x→0,y→0 x→0,y→0
1.画出方程表示的曲面：z= -（√(x^2+y^2)）2.证明极限lim [(x+y)/(x-y)]不存在x→0,y→03.求函数极限lim[(x+y)sin(1/x^2+y^2)],lim[(xy)/（√（xy+1））-1]x→0,y→0 x→0,y→0
1.这是一个顶点在坐标原点圆锥体曲面（画图形有点麻烦,约去）；
2.∵设y=kx
代入得lim(x->0,y->0) [(x+y)/(x-y)]=lim(x->0,y->0)[(1+k)/(1-k)]=(1+k)/(1-k)
显然,k取不同的值时,所求极限有不同的结果
∴极限lim(x->0,y->0) [(x+y)/(x-y)]不存在；
3.(1)∵sin(1/x^2+y^2)是有界函数
∴lim(x->0,y->0)sin(1/x^2+y^2)=M(有限值)
∵lim(x->0,y->0)(x+y)=0
∴lim(x->0,y->0)[(x+y)sin(1/x^2+y^2)]=[lim(x->0,y->0)(x+y)]*[lim(x->0,y->0)sin(1/x^2+y^2)]
=0*M
=0；
(2)lim(x->0,y->0)[(xy)/(√(xy+1)-1)]=lim(x->0,y->0)[(xy)(√(xy+1)+1)/(xy+1-1) (分子分母同乘以√(xy+1)+1)
=lim(x->0,y->0)[√(xy+1)+1]
=√(0+1)+1
=2.