This exercise is asking me to study the continuity of the following function: $$f(x, y) = \frac{\ln (1+x^4-y^4)}{(x^2-y^2)^2}$$ I check the domain of the function: $$1+x^4-y^4 > 0$$ and the important part is that we have to exclude those points from the $\;y = x\;$ and $\;y = -x\;$ lines. For $\;y = x\;$ I followed this approach:
$$\begin{align}\lim_{y \to x^+}\frac{\ln (1+x^4-y^4)}{(x^2-y^2)^2} \approx \\\lim_{y \to x^+}\frac{x^4-y^4}{(x^2-y^2)^2} = \\\lim_{y \to x^+}\frac{(x^2+y^2)(x^2-y^2)}{(x^2-y^2)^2} = \\\lim_{y \to x^+}\frac{x^2+y^2}{x^2-y^2} = -\infty\end{align}$$ Also: $$\lim_{y \to x^-}\frac{x^2+y^2}{x^2-y^2} = +\infty$$ Same would work with $\;y = -x.\;$ I wonder if this is correct as a friend told me I can't do this because either one of the variables must be constant. Anyway, I'm looking for a way to show that the function goes to infinity when $\;(x, y)\;$ goes near the $\;y = x\;$ and $\;y = -x.$ Another approach I've come with is to cut, for example, the 3d graph with the plane $y = x,$ orthogonal to the $y = -x$ one, and then I would find the intersection, and therefore a single variable graph which would go to infinity at... $x = 0?$ The problem with this approach is that I have really no idea how to do it. It seems like I can't just substitute $y = x$ in the function, whatever that means. I have been looking for a similar question in the forum but all I've found were questions regarding single point discontinuities, and my question involves straight lines. Any other approach would be really helpful. This was my first question here, I hope I was clear enough. Thank you.
After having thought about this for a while, I havent't quite figured it out, however I have a suggestion. If you factor $(x^2 - y^2)^2$ as $(x+y)^2 (x-y)^2$ then you can rewrite the original expression and have one factor that does not evaluate to $0$ when taking the limit as $x$ approaches $y$. Now you are left with an indeterminate limit, so you can use L'Hopital's rule after this. All of this should yield an expression you can continue to manipulate, since you have gotten rid of the squared terms inside the paranthesis in the numerator in the original expression. This is as far as I've tried to work the problem. It seems promising though, so it might be worth a shot. I tried plotting the graph and it looks a little weird, so you might not have a well defined limit at $x=y$, meaning you should consider the limit when $x$ goes to $y$ from above and when $x$ goes to $y$ from below.
Hope this gets you somewhere if you decide to try it out! Best of luck.