Thomas' Calculus 13th Edition

Consider first approach : $(x,y) \to (0,0)$ along $y=0$ This implies that $\lim\limits_{x \to 0}\dfrac{x^4}{x^4+(0)^2}=1$ Next, let us consider our second approach : $(x,y) \to (0,0)$ along $y=x^2$ This implies that $\lim\limits_{x \to 0}\dfrac{x^4}{x^4+(x^2)^2}=\dfrac{1}{2}$ This shows that there are different limit values for different approach, so, the limit does not exist at the point (0,0) for the function $f(x,y)=\dfrac{x^4}{x^4+y^2}$.