本话题讨论多元函数可微性判断的相关习题, 欢迎大家参与讨论!
判断函数 z=f\left(x,y\right) 在点 (x_0, y_0) 处是否可微, 步骤如下:
(1) 写出全增量 \Delta z=f\left(x_0+\Delta x,y_0+\Delta y\right)-f\left(x_0,y_0\right);
(2) 写出线性增量 A\Delta x+B\Delta y,其中 A=f_x^{\prime}(x_0,y_0),B=f_y^{\prime}(x_0,y_0);
(3) 作极限 \lim\limits_{\Delta x \to 0 \atop \Delta y \to 0} \frac{\Delta z-(A\Delta x+B\Delta y)}{\sqrt{\left(\Delta x\right)^{2}+\left(\Delta y\right)^{2}}}, 若该极限等于0, 则 z=f(x,y) 在点 (x_0,y_0) 处可微, 否则, 就不可微.
典型习题
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设函数 f(x, y) = \left\{ \begin{array}{ll} \frac{xy}{\sqrt{x^2 + y^2}}, & x^2 + y^2 \neq 0 \\ 0, & x^2 + y^2 = 0 \end{array} \right. , 则 f(x,y) 在点(0,0)处是否可微?
解：不可微。
f_{x}^{\prime}(0,0)=\lim\limits_{\Delta x\to0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=0.
f_{y}^{\prime}(0,0)=\lim\limits_{\Delta y\rightarrow0}\frac{f(0,0+\Delta y)-f(0,0)}{\Delta y}=0.
\Delta z=f(0+\Delta x,0+\Delta y)-f(0,0)= \frac{\Delta x\cdot\Delta y}{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}}
\lim\limits_{\Delta x\to0,\Delta y\to0}\frac{\Delta z-[f_{x}^\prime(0,0)\cdot\Delta x+f_{y}^\prime(0,0)\cdot\Delta y]}{\sqrt{{(\Delta x)^{2}+(\Delta y)^{2}}}}=\lim\limits_{\Delta x\to0,\Delta y\to0}\frac{\Delta x\cdot\Delta y}{(\Delta x)^{2}+(\Delta y)^{2}}
令 x=k y, 得
\lim\limits_{\Delta x\to0,\Delta y\to0}\frac{\Delta x\cdot\Delta y}{(\Delta x)^{2}+(\Delta y)^{2}}=\lim\limits_{\Delta x\to0,\Delta x\to0}\frac{\Delta x\cdot k\Delta x}{(\Delta x)^{2}+(k\Delta x)^{2}}=\frac{k}{1+k^2}, 极限不存在.
所以, f(x,y) 在（0,0）点处不可微. -
已知
f(x) = \begin{cases} \frac{\sqrt{\lvert xy \rvert}}{x^2 + y^2} \sin(x^2 + y^2), & (x,y) \neq (0,0), \\ 0, & (x,y) = (0,0), \end{cases}
判断 f(x) 在(0,0)处偏导数存在性以及可微性.
解：
(1)f_x^{\prime}(0,0)=\lim\limits_{\Delta x\to0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=0;
f_y^{\prime}(0,0)=\lim\limits_{\Delta y\rightarrow0}\frac{f(0,0+\Delta y)-f(0,0)}{\Delta y}=0.
所以, f(x,y) 在（0,0）点处偏导数存在.
(2) \lim\limits_{(x,y)\rightarrow(0,0)}\frac{\frac{\sqrt{|xy|}}{x^2+y^2}\sin(x^2+y^2)-f_x(0,0)x-f_y(0,0)y}{\sqrt{x^2+y^2}}=\lim\limits_{(x,y)\rightarrow(0,0)}\sqrt{\frac{|xy|}{x^2+y^2}}
当 x=ky 时,
\lim\limits_{(x,y)\rightarrow(0,0)}\sqrt{\frac{|xy|}{x^2+y^2}}=\lim\limits_{(x,y)\rightarrow(0,0)}\sqrt{\frac{|k|}{k^2+1}} 极限不存在,
故 f(x,y) 在 (0,0) 处不可微. -
设 f(x, y) = \begin{cases} (x^2 + y^2) \sin\frac{1}{x^2 + y^2}, & x^2 + y^2 \neq 0, \\ 0, & x^2 + y^2 = 0.\end{cases}
则 f(x,y) 在点(0,0)处是否可微?
解：可微.
f_{x}^{\prime}(0,0)=\lim\limits_{\Delta x\to0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x}=\lim\limits_{\Delta x\to0}\frac{(\Delta x)^{2}\sin\frac{1}{(\Delta x)^{2}}}{\Delta x}=0,
f_y^{\prime}(0,0)=\lim\limits_{\Delta y\to0}\frac{f(0,0+\Delta y)-f(0,0)}{\Delta y}=\lim\limits_{\Delta y\to0}\frac{(\Delta y)^{2}\sin\frac{1}{(\Delta y)^{2}}}{\Delta y}=0.
\Delta z=f(0+\Delta x,0+\Delta y)-f(0,0)=[(\Delta x)^{2}+(\Delta y)^{2}]\cdot\sin\frac{1}{(\Delta x)^{2}+(\Delta y)^{2}}.
\begin{aligned} & \lim\limits_{\Delta x\to0,\Delta y\to0}\frac{\Delta z-[f_{x}^\prime(0,0)\cdot\Delta x+f_{y}^\prime(0,0)\cdot\Delta y]}{\sqrt{{(\Delta x)^{2}+(\Delta y)^{2}}}} \\ &=\lim\limits_{\Delta x\to0,\Delta y\to0}\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}\cdot\sin\frac{1}{(\Delta x)^{2}+\Delta y)^{2}}. \end{aligned}
令 \rho=\sqrt{{(\Delta x)^{2}+(\Delta y)^{2}}}, 则
\lim\limits_{\Delta x\to0,\Delta y\to0}\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}\cdot\sin\frac{1}{(\Delta x)^{2}+\Delta y)^{2}}=\lim\limits_{ \rho\to0}\rho\cdot\sin\frac{1}{\rho^2}=0.
所以, f(x,y) 在（0,0）点处可微.