Does one always need to prove that $\left|f(x,y)-L\right|<\epsilon$ for a limit if the point itself does not exist? I left another question on how to get to the point of proving that for $\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2}$ for $a\ne 0$ however I got the comment that it wasn't neccesary to prove that $\lim^{}_{(x,y) \rightarrow (a,a)} \frac{x^3-y^3 }{x^2-y^2} < \delta$. What do I need to do here?
(previous question: https://math.stackexchange.com/posts/2145811)
It's totally unclear what you mean.
The expression at stake is undefined on the two lines $x=\pm y$. But whenever $x\ne y$ one has $$f(x,y):={x^3-y^3\over x^2-y^2}={(x-y)(x^2+xy+y^2)\over (x-y)(x+y)}={x^2+xy+y^2\over x+y}\ ,$$ and the right hand side is continuous on the set $C:=\bigl\{(x,y)\in{\mathbb R}^2\,\bigm|\,x+y\ne0\bigr\}$, in particular at the point $(a,a)$ when $a\ne0$.
If we extend the given $f$ continuously to the points of $C$ (a voluntary act not enforced by the problem statement) then we obtain the function $$\tilde f(x,y):={x^2+xy+y^2\over x+y}\qquad\bigl((x,y)\in C\bigr)\ ,$$ and $$\lim_{(x,y)\to(a,a)}\tilde f(x,y)=\tilde f(a,a)={3a\over2}\qquad(a\ne0)\ .$$