证明已知f(x,y)=(x3+y3)/(x2+y),求f(x,y)在点（0,0）处的两个二次极限并证明重极限不存在

Question

证明已知f(x,y)=(x3+y3)/(x2+y),求f(x,y)在点（0,0）处的两个二次极限并证明重极限不存在

Answer 1

lim[x→0]lim[y→0] (x³+y³)/(x²+y)
=lim[x→0] x³/x²
=lim[x→0] x
=0
lim[y→0]lim[x→0] (x³+y³)/(x²+y)
=lim[y→0] y³/y
=lim[y→0] y²
=0
下面证明二重极限不存在
令(x,y)沿y=x趋于(0,0)
lim[x→0,y→0] (x³+y³)/(x²+y)
=lim[x→0] 2x³/(x²+x)
=lim[x→0] 2x²/(x+1)
=0
令(x,y)沿y=x³-x² 趋于(0,0)
lim[x→0,y→0] (x³+y³)/(x²+y)
=lim[x→0] [x³+(x³-x²)³]/(x²+x³-x²)
=lim[x→0] [x³+(x³-x²)³]/x³
=lim[x→0] [1+(x³-x²)³/x³]
=1
因此极限不存在.
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