## Thomas' Calculus 13th Edition

The path limit along $x=0$ and $y=kx$ is equal to $0$.
We are given that $f(x,y)=\dfrac{2x^2y}{x^4+y^2}$ a) When $x=0$ Then, we have $\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=0$ b) Next, we have the second approach along $y=kx$ So, $\lim\limits_{(x,y) \to (0,0)}\dfrac{2x^2y}{x^4+y^2}=\dfrac{2x^2(kx)}{x^4+(kx)^2}=0$ This implies that the path limit along $x=0$ and $y=kx$ is equal to $0$.