x→0,y→0,A = [lim xln(1+xy)]/(x+y) = [lim (x*xy)]/(x+y) = lim[(x^2*y)/(x+y)]
f(x,y) = (x^2*y)/(x+y)
x→0,y=kx ,limf(x,y) =lim(k*x^3)/(x+kx) = lim[(k/(1+k)]*x^2 = 0
由此假设:x→0,y→0,limf(x,y) = 0
试用定义证明
|f(x,y) - 0| = | (x^2*y)/(x+y) - 0 |
=x^2* |y| / |x+y|
<= (x^2+y^2) *|y| /|x+y|
<= x^2+y^2
可见任意a>0,取b=(a)^(1/2) ,则当
0<[(x-0)^2 + (y-0)^2] ^(1/2) < b
即P(x,y)属于f(x,y)定义域内(0,0)的去心邻域时,总有 |f(x,y) - 0| < a
成立,所以 x→0,y→0 ,limf(x,y) = 0
即x→0,y→0,[lim xln(1+xy)]/(x+y) = 0
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