# Finding $\lim \limits_{(x, y)\rightarrow (1, -1)}\frac{(x-y)^2-4}{x^2+y^2-2}$

I want to calculate $\lim\limits_{(x, y)\rightarrow (1, -1)}\dfrac{(x-y)^2-4}{x^2+y^2-2}$.

I already know that $\lim \limits_{x\rightarrow 1}\lim\limits_{y\rightarrow -1}\dfrac{(x-y)^2-4}{x^2+y^2-2} = \lim\limits_{y\rightarrow -1}\lim\limits_{x\rightarrow 1}\dfrac{(x-y)^2-4}{x^2+y^2-2} = 2$, so if the limit exists, it equals $2$. But how do I prove that?

I can't use polar coordinates since $(x, y) \not\to (0,0)$, I've also failed to find a counterexample of a sequence with different limit than $2$.

• Consider a change of coordinates so that the limit goes to (0, 0). Then it should be easier to compute it. – RGS Nov 8 '16 at 22:36

Note that for all $x,y$ such that $x^2+y^2\neq 2$ the following holds:$$\frac{(x-y)^2-4}{x^2+y^2-2}=\dfrac{x^2+y^2-2-2xy-2}{x^2+y^2-2}=1-\dfrac{2xy+2}{x^2+y^2-2}.$$
Conjecture: $\lim \limits_{(x,y)\to (1,-1)}\left(\dfrac{2xy+2}{x^2+y^2-2}\right)=-1$.
Note that $$\dfrac{2xy+2}{x^2+y^2-2}+1=\dfrac{2xy+2}{x^2+y^2-2}+\dfrac{x^2+y^2-2}{x^2+y^2-2}=\dfrac{(x+y)^2}{x^2+y^2-2},$$
therefore $$\left|\dfrac{2xy+2}{x^2+y^2-2}+1\right|=\left|\dfrac{(x+y)^2}{x^2+y^2-2}\right|\leq \left|\dfrac{(x+y)^2}{x^2+y^2}\right|_.$$