## Calculus: Early Transcendentals 8th Edition

We notice that if we directly substitute limits in the given function $f(x,y)=\frac{xy-y}{(x-1)^{2}+y^{2}}$ Then $f(0,0)=\frac{0}{0}$ To evaluate limit along x-axis; put $y=0$ $f(x,0)=\frac{xy-y}{(x-1)^{2}+y^{2}}=\frac{x.0-0}{(x-1)^{2}+y^{2}}=0$ To evaluate limit along y-axis; put $x=1$ $f(1,y)=\frac{xy-y}{(x-1)^{2}+y^{2}}=\frac{y-y}{(1-1)^{2}+y^{2}}=0$ However, the obtained identical limits along the axes do not show that the given limit is 0. Then, approach (0,0) along another line,$y=x-1$ and $x\ne0$ $f(x,x-1)=\frac{x(x-1)-(x-1)}{(x-1)^{2}+(x-1)^{2}}=\frac{1}{2}$ For a limit to exist, all the paths must converge to the same point. Since $f(x,y)$ has two different values along two different paths, it follows that the limit does not exist.