How to find $\lim_{(x,y)\to(1,1)} \frac{x^3-y^3}{x^2 - y^2}$? Suppose $$ f(x,y) = \begin{cases}
\frac{x^3-y^3}{x^2 - y^2},  & \text{if $x^2 \ne y^2$} \\
\alpha, & \text{if $x^2 = y^2$}  \\
\end{cases}$$
I need to find $\alpha$ such that $f(x,y)$ will be continous at $(1,1)$.
I started:
If $f(x,y)$ continous at $(1,1)$ $\implies$ $\lim_{(x,y)\to(1,1)} \frac{x^3-y^3}{x^2 - y^2} = \alpha$
Then I need to find $\lim_{(x,y)\to(1,1)} \frac{x^3-y^3}{x^2 - y^2}$ but I stuck here.
Can you please help me find the limit? I'd glad to a detailed solution or hint .. thanks in advance..

Hint: Notice that the numerator factors as follows:
$$
x^3 - y^3 = (x - y)(x^2 + xy + y^2)
$$
Now factor the denominator and cancel the $(x-y)$ factor.
