本文通过一个具体的数学例子，展示了函数可偏导但不连续的情况，即函数f(x,y)在点(0,0)处虽然存在偏导数，但却不是连续的。
可偏导不一定连续的例子
f(x,y)={xyx2+y2,(x,y)≠(0,0) 0,(x,y)=(0,0)}(1)f(x,y)=\left\{\begin{array}{l} \frac{xy}{x^2+y^2}&, &(x,y)\ne(0,0)\\ \ \ \ \ 0&,&(x,y)=(0,0)\end{array} \right\}\tag{1}f(x,y)={x2+y2xy 0,,(x,y)=(0,0)(x,y)=(0,0)}(1)
可偏导的证明
fx(0,0)=limΔx→0f(0+Δx,0)−f(0,0)Δx=limΔx→00(Δx)2−0Δx=0(2)\begin{aligned} f_x(0,0)& = \lim\limits_{\Delta x \to 0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}\\ & =\lim\limits_{\Delta x \to 0}\frac{\frac{0}{(\Delta x)^2}-0}{\Delta x}\\ & =0 \tag{2} \end{aligned}fx(0,0)=Δx→0limΔxf(0+Δx,0)−f(0,0)=Δx→0limΔx(Δx)20−0=0(2)
同理，fy(0,0)=0f_y(0,0)=0fy(0,0)=0.所以，f(x,y)f(x,y)f(x,y)存在偏导数
不连续的证明
limx→0t→kxf(x,y)=limx→0kx2x2+k2x2=k1+k2(3)\begin{aligned} \lim\limits_{\tiny\begin{array}{l}x\to0\\t\to kx \end{array}}f(x,y)&= \lim\limits_{\tiny x\to 0}\frac{kx^2}{x^2+k^2x^2}\\ &=\frac{k}{1+k^2}\tag{3} \end{aligned}x→0t→kxlimf(x,y)=x→0limx2+k2x2kx2=1+k2k(3)
所以f(x,y)f(x,y)f(x,y)在(0,0)(0,0)(0,0)处不连续.
扫一扫
2747
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
