下列函数在点$(0,0)$的重极限存在的是
A: $f(x,y)=\frac{y^2}{x^2+y^2}$
B: $f(x,y)=(x+y)\sin\frac{1}{x}\sin\frac{1}{y}$
C: $f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$
D: $f(x,y)=\frac{x^2y^2}{x^3+y^3}$
A: $f(x,y)=\frac{y^2}{x^2+y^2}$
B: $f(x,y)=(x+y)\sin\frac{1}{x}\sin\frac{1}{y}$
C: $f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$
D: $f(x,y)=\frac{x^2y^2}{x^3+y^3}$
举一反三
- 4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$，并且$z(x,0)=x,z(0,y)={{y}^{2}}$，则$z(x,y)=$（ ） A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$
- 函数$f(x,y)={{\text{e}}^{-x}}\cos y$在点$(0,0)$处2次Taylor多项式为 A: $1+x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ B: $1-x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$ C: $1-x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$ D: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$
- $(0,0)$是以下函数$f(x,y)$的定义域的内点是 A: $f(x,y)=\sqrt{x} \ln(x+y)$ B: $f(x,y)=\frac{x+y}{x^2+y^2 }$ C: $f(x,y)=\arcsin \frac{x}{y}$ D: $f(x,y)=\ln (1-x^2-y^2)$
- 函数$f(x,y)=\sin x\cdot \ln (1+y)$在点$(0,0)$处带有Peano型余项的3阶Taylor公式为$f(x,y)=$ A: $xy+\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ B: $xy-\frac{1}{2}x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ C: $xy-x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$ D: $xy+x{{y}^{2}}+o({{(\sqrt{{{x}^{2}}+{{y}^{2}}})}^{3}})$
- 方程$(x^2+1)(y^2-1) + xy y' = 0$的通解为 A: $y^2 = C \frac{e^{-x^2}}{x^2}$ B: $y = C \frac{e^{-x^2}}{x^2}$ C: $y^2 = C \frac{e^{-x^2}}{x^2}+1$ D: $y=C \frac{e^{-x^2}}{x^2}+1$