Prove that the limit doesn't exists I have to prove that
$$\lim_{(x,y) \to (0,0)} \frac{(x+y)^3}{(\sqrt{x^2-y^2})^2}$$
doesn't exist.
I think I have tried every way possible to show that doesn't exist $(x=0;y=0; x=my; x=my^2;...)$, but nothing works. I get 0 every time.
Does anyone have any suggestions?
Thank you

What you can do is to find the limit along a path $y = x\gamma(x)$. Using that the expression becomes:
$${(x+y)^3\over\left(\sqrt{x^2-y^2}\right)} ={(x+x\gamma(x))^3\over\left(\sqrt{x^2-x^2\gamma(x)^2}\right)^2} = {x^3(1+\gamma(x))^3\over x^2(1-\gamma(x)^2)} = {x (1+\gamma(x))^2\over 1-\gamma(x)}$$
Now if we select $\gamma(x)$ such that $1-\gamma(x) = x/C$ (ie $\gamma(x) = 1-x/C$) we get
$${(x+y)^3\over\left(\sqrt{x^2-y^2}\right)^2}={x(1+1-x/C)^2\over 1-(1-x/C)} = {x(2-x/C)^2\over x/C}=C(2-x/C)^2 \to 4C$$
Now we have constructed paths to origin with arbitrary limit, but for the limit to exist the path-wise limit must be independent of the path we chose.
