Question

The function f(x,y)={x^2y/x^4+y^2 (x,y) is not=0 0, (x,y)=0 is not continuous at (0,0)

Answer 1

If $x=0$ then $f(0, y)=\dfrac{0(y)}{0+y^2}=0.$ Therefore

$$f(x, y)\to 0\text{ as} \ (x,y)\to(0,0) \text{ along the }y-\text{axis}$$

For all $x\not=0$

$$f(x, x^2)=\dfrac{x^2(x^2)}{x^4+(x^2)^2}=\dfrac{1}{2}\not=0$$

$$f(x, y)\to \dfrac{1}{2}\not=0\text{ as} \ (x,y)\to(0,0) \text{ along }y=x^2$$

Since we have obtained different limits along different paths, limit

$$\lim\limits_{(x,y)\to(0,0)}f(x, y)=\lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y}{x^4+y^2}$$

does not exist.

The function $f(x, y)=\dfrac{x^2y}{x^4+y^2}$ is not continuous at $(0,0).$