## Thomas' Calculus 13th Edition

Consider our approach : $(x,y) \to (0,0)$ along $y=kx; m\ne -10$ Then, we get $\lim\limits_{x \to 0} \dfrac{x-kx}{x+kx}=\lim\limits_{x \to 0} \dfrac{x(1-k)}{x(1+k)}=\dfrac{1-k}{1+k}$ 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 function $f(x,y)=\dfrac{x-y}{x+y}$.