Whats is the limit of $\lim_{(x,y)\to(a,a)}\frac{x^3-y^3}{x^2-y^2}$ for $a\ne0$? What is $\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}$ for $a\ne 0$? I have found that the limit coould be $\frac{3a}{2}$ to proof it I have come to the point where $\left|\frac{x^3-y^3 }{x^2-y^2}-\frac{3a}{2}\right| = \left|\frac{x^2+xy+y^2 }{x+y}-\frac{3a}{2}\right|\le \left|x+y-\frac{3a}{2}\right|$. I have to go to something like $\sqrt{(x-a)^2+(y-a)^2}$ however I have no idea what to do from here.

$$\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{(x-y)(x^2+xy+y^2) }{(x-y)(x+y)}=\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^2+xy+y^2}{x+y}=\frac{3a}{2}$$
