多元函数的极限、连续与偏导数

1，极限：多元函数的极限

$\lim_{(x,y)\to(a,b)}f(x,y)=A$

$\lim_{{(x,y)\to (0,0)}\atop{y=x}}\frac{xy}{x^2+2y^2}=\lim_{{x\to 0}\atop{y=x}}\frac{x^2}{x^2+2x^2}=\frac{1}{3}$

$\lim_{{(x,y)\to (0,0)}\atop{y=0}}\frac{xy}{x^2+2y^2}=0$

$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{5xy}=\lim_{u\to 0}\frac{\sin u}{5u}=\frac{1}{5}$

$\left|\frac{x^2y^2}{x^2+y^2}\right|\le \left|\frac{x^2y^2}{x^2}\right|=y^2\to 0$

2，连续：若 $\lim_{(x,y)\to(a,b)}f(x,y)=f(a,b)$

3，偏导数：设 $$z=f(x,y)$$，则它在点 $$(a,b)$$ 处对 $$x$$ 的偏导数定义为

$\frac{\partial z}{\partial x}(a,b)=\lim_{x\to a}\frac{f(x,b)-f(a,b)}{x-a}$

$\frac{\partial z}{\partial y}(a,b)=\lim_{y\to b}\frac{f(a,y)-f(a,b)}{y-b}$

$\frac{\partial z}{\partial x}=\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}$

$\frac{\partial z}{\partial y}=\lim_{h\to 0}\frac{f(x,y+h)-f(x,y)}{h}$

\begin{align*}\frac{\partial z}{\partial y}&=2ye^{x^2+y^2}\sin(xy)+xe^{x^2+y^2}\cos(xy)\end{align*}

4，高阶偏导数：对一阶偏导数再求偏导数就是二阶偏导数，依此类推。

$\frac{\partial z}{\partial x},\quad \frac{\partial z}{\partial y}$

$\frac{\partial^2z}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial x}\right)=f_{xx},\quad\frac{\partial^2z}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial x}\right)=f_{xy}$

$\frac{\partial^2z}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial z}{\partial y}\right)=f_{yx},\quad\frac{\partial^2z}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{\partial z}{\partial y}\right)=f_{yy}$