关于二元分段函数求极限
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关于二元分段函数求极限
consider the function f:R^2 --〉R given by f(x.y)=0 (if x=0,y>=0) or=-x(if x>=0.y
consider the function f:R^2 --〉R given by f(x.y)=0 (if x=0,y>=0) or=-x(if x>=0.y
![关于二元分段函数求极限](/uploads/image/z/17793500-68-0.jpg?t=%E5%85%B3%E4%BA%8E%E4%BA%8C%E5%85%83%E5%88%86%E6%AE%B5%E5%87%BD%E6%95%B0%E6%B1%82%E6%9E%81%E9%99%90)
f(x.y) =0 (x=0,y>=0) in the 1st quadrant
=-x (x>=0,y(0-,0-) limit f(x,y) = 0
(x,y)->(0+,0+) limit f(x,y) = 0
(x,y)->(0-,0+) limit f(x,y) = 0
(x,y)->(0+,0-) limit f(x,y) = 0
f(0,0)=0
Therefore,the function f(x.y) is continuous at the point (0,0),and all the other points in the region of R².
∂f/∂x = 0 (x=0,y>=0)
= -1 (x>=0,y=0)
= 0 (x>=0,y
=-x (x>=0,y(0-,0-) limit f(x,y) = 0
(x,y)->(0+,0+) limit f(x,y) = 0
(x,y)->(0-,0+) limit f(x,y) = 0
(x,y)->(0+,0-) limit f(x,y) = 0
f(0,0)=0
Therefore,the function f(x.y) is continuous at the point (0,0),and all the other points in the region of R².
∂f/∂x = 0 (x=0,y>=0)
= -1 (x>=0,y=0)
= 0 (x>=0,y