请教多元函数的极限 可导性 连续性的问题解法 俩题~~

Question

如题所述

Answer 1

① 由均值不等式, -(x²+y²)/2 ≤ xy ≤ (x²+y²)/2,
故|f(x,y)| ≤ |(x²+y²)/2|/√(x²+y²) = √(x²+y²)/2 → 0, 当(x,y) → (0,0).
于是lim{(x,y) → (0,0)} f(x,y) = 0 = f(0,0).
即在原点极限存在且连续.

在原点, ∂f/∂x = lim{x → 0} (f(x,0)-f(0,0))/x = 0,
∂f/∂y = lim{y → 0} (f(0,y)-f(0,0))/x = 0, 即两个偏导数存在并得0.
但沿y = x方向的方向导数lim{x → 0} (f(x,x)-f(0,0))/√2x = 1/2 ≠ 0.
故f(x,y)在原点不是可微的.

② 当x = 0时, 有f(x,y) = 0, 故(x,y)沿x = 0趋近原点时, f(x,y) → 0.
而当x = y时, 有f(x,y) = 1, 故(x,y)沿y = x趋近原点时, f(x,y) → 1.
因此lim{(x,y) → (0,0)} f(x,y)不存在.
于是f(x,y)在原点不连续, 也不可微.

在原点, ∂f/∂x = lim{x → 0} (f(x,0)-f(0,0))/x = 0,
∂f/∂y = lim{y → 0} (f(0,y)-f(0,0))/x = 0, 即两个偏导数存在并得0.