第6章 6.19

f(x,y)={x+xyx2+y2(x,y)≠(0,0)0(x,y)=(0,0)f_{(x,y)} = \begin{cases} x + \dfrac{xy}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}

fx=1+y(x2+y2)−xy⋅2x(x2+y2)2=1+−x2y+y3(x2+y2)2f_x = 1 + \frac{y(x^2+y^2) - xy \cdot 2x}{(x^2+y^2)^2} = 1 + \frac{-x^2y+y^3}{(x^2+y^2)^2}

lim⁡(x,y)→(0,0)fx=1\lim_{(x,y)\to(0,0)} f_x = 1

fy=x(x2+y2)−xy⋅2y(x2+y2)2=x3−xy2(x2+y2)2f_y = \frac{x(x^2+y^2) - xy \cdot 2y}{(x^2+y^2)^2} = \frac{x^3-xy^2}{(x^2+y^2)^2}

lim⁡(x,y)→(0,0)fy=0\lim_{(x,y)\to(0,0)} f_y = 0

よって原点で連続でない

解説: r31bn1z