## Calculus 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 $f(0,0)=\frac{0}{0}$ We get intermediate form. Therefore, we will calculate limit of function in following way. 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 $x=1$ $f(1,y)=\frac{xy-y}{(x-1)^{2}+y^{2}}=\frac{y-y}{(1-1)^{2}+y^{2}}=0$ Although the obtained identical limits along the axes do not show that the given limit is 0. Then, approach (0,0) along another line let us say $y=x-1$ 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, function $f(x,y)$ has two different values along two different paths, it follows that limit does not exist. Hence, both the limits are different and follow different paths, therefore, the limits does not exist.