例 1.
令 $y=kx$,
\[
\lim_{x\rightarrow 0}f(x,kx)=\lim_{x\rightarrow 0}\frac{x\cdot kx}{x^2+(kx)^2}=\frac{k}{1+k^2}.
\]
对于不同的 $k$, 取值不同, 因此极限不存在. 故在 $(0,0)$ 点不连续.
不过在 $(0,0)$ 点处的偏导数却是存在的.
\[
\frac{\partial f}{\partial x}(0,0)=\frac{d}{dx}f(x,0)\bigr|_{x=0}=\lim_{x\rightarrow 0}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\rightarrow 0}\frac{\frac{x\cdot 0}{x^2+0^2}-0}{x-0}=0.
\]
\[
\frac{\partial f}{\partial y}(0,0)=\frac{d}{dy}f(0,y)\bigr|_{y=0}=\lim_{y\rightarrow 0}\frac{f(0,y)-f(0,0)}{y-0}=\lim_{y\rightarrow 0}\frac{\frac{0\cdot y}{0^2+y^2}-0}{y-0}=0.
\]
例 3.
函数在 $(0,0)$ 处是连续的. 因为
\[
\biggl|xy\frac{x^2-y^2}{x^2+y^2}\biggr|\leqslant\frac{x^2+y^2}{2}\cdot\biggl|\frac{x^2-y^2}{x^2+y^2}\biggr|=\frac{|x^2-y^2|}{2},
\]
而 $\lim\limits_{(x,y)\rightarrow(0,0)}|x^2-y^2|=0$, 故 $\lim\limits_{(x,y)\rightarrow(0,0)}xy\frac{x^2-y^2}{x^2+y^2}=0$.
\[
\frac{\partial f}{\partial x}(0,0)=\frac{d}{dx}f(x,0)\bigr|_{x=0}=\lim_{x\rightarrow 0}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\rightarrow 0}\frac{x\cdot 0\frac{x^2-0^2}{x^2+0^2}-0}{x-0}=0.
\]
\[
\frac{\partial f}{\partial y}(0,0)=\frac{d}{dy}f(0,y)\bigr|_{y=0}=\lim_{y\rightarrow 0}\frac{f(0,y)-f(0,0)}{y-0}=\lim_{y\rightarrow 0}\frac{0\cdot y\frac{0^2-y^2}{0^2+y^2}-0}{y-0}=0.
\]
当 $p=(x,y)\neq(0,0)$ 时,
\[
\begin{split}
\frac{\partial f}{\partial x}\bigr|_{p}&=\frac{\partial}{\partial x}(xy\frac{x^2-y^2}{x^2+y^2})=y\cdot\frac{x^2-y^2}{x^2+y^2}+xy\frac{2x(x^2+y^2)-(x^2-y^2)\cdot 2x}{(x^2+y^2)^2}\\
&=y\cdot\frac{x^2-y^2}{x^2+y^2}+xy\cdot\frac{4xy^2}{(x^2+y^2)^2}\\
&=y\frac{(x^2-y^2)(x^2+y^2)+4x^2y^2}{(x^2+y^2)^2}=y\cdot\frac{x^4+4x^2y^2-y^4}{(x^2+y^2)^2},
\end{split}
\]
所以 $f'_x(0,y)=y\cdot\frac{-y^4}{y^4}=-y$.
\[
\begin{split}
\frac{\partial f}{\partial x}\bigr|_{p}&=\frac{\partial}{\partial y}(xy\frac{x^2-y^2}{x^2+y^2})=x\cdot\frac{x^2-y^2}{x^2+y^2}+xy\frac{-2y(x^2+y^2)-(x^2-y^2)\cdot 2y}{(x^2+y^2)^2}\\
&=x\cdot\frac{x^2-y^2}{x^2+y^2}+xy\cdot\frac{-4x^2 y}{(x^2+y^2)^2}\\
&=x\frac{(x^2-y^2)(x^2+y^2)-4x^2 y^2}{(x^2+y^2)^2}=x\cdot\frac{x^4-4x^2 y^2-y^4}{(x^2+y^2)^2},
\end{split}
\]
所以 $f'_y(x,0)=x\cdot\frac{x^4}{x^4}=x$.
\[
f''_{xy}(0,0)=\lim_{y\rightarrow 0}\frac{f'_x(0,y)-f'_x(0,0)}{y-0}=\lim_{y\rightarrow 0}\frac{-y-0}{y-0}=-1.
\]
\[
f''_{yx}(0,0)=\lim_{x\rightarrow 0}\frac{f'_y(x,0)-f'_y(0,0)}{x-0}=\lim_{x\rightarrow 0}\frac{x-0}{x-0}=1.
\]
由此可见, $f''_{xy}$ 和 $f''_{yx}$ 在 $(0,0)$ 处不连续. (参见定理)
