## Thomas' Calculus 13th Edition

Consider a approach : $(x,y) \to (0,0)$ along $y=mx; m\ne 1$ Then, we get $\lim\limits_{x \to 0}\dfrac{x^2-kx}{x-kx}=\lim\limits_{x \to 0}\dfrac{x(x-k)}{x(1-k)}=\dfrac{-k}{1-k}$ This shows that there are multiple limit values when the approach is different, so, the limit does not exist at the point (0,0) for the given function $f(x,y)=\dfrac{x^2-y}{x-y}$.