If x=0 then f(0,y)=0+3y20(y)=0. Therefore
f(x,y)→0 as (x,y)→(0,0) along the y−axis For all x=0
f(x,x5)=x10+3(x5)24x5(x5)=1=0
f(x,y)→1=0 as (x,y)→(0,0) along y=x5Since we have obtained different limits along different paths, limit
(x,y)→(0,0)limf(x,y)=(x,y)→(0,0)limx10+3y24x5y
does not exist.
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