What is $\lim_{(x,y)\to(0,0)} \frac{(x^3+y^3)}{(x^2-y^2)}$? In class, we were simply given that this limit is undefined since along the paths $y=\pm x$, the function is undefined.
Am I right to think that this should be the case for any function, where the denominator is $x^2-y^2$, regardless of what the numerator is?
Just wanted to see if this is a quick way to identify limits of this form.
Thanks for the discussion and help!

If you define $$\lim_{\langle x,y\rangle\to\langle a,b\rangle}f(x,y)\tag{1}$$ in such a way that it exists only when the function is defined in some open ball centred at $\langle a,b\rangle$, then what you wrote is correct. This is analogous to defining $$\lim_{x\to a}f(x)$$ only when $f(x)$ is defined in some open interval centred at $a$. However, just as we can talk about one-sided limits on the real line, it makes perfectly good sense to talk about $(1)$ whenever $f(x,y)$ is defined at points arbitrarily close to $\langle a,b\rangle$. In that case it’s understood that we look only at the limit along ‘paths’ within the domain of $f$. On that understanding $$\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{x^3+y^3}{x^2-y^2}$$ still does not exist, but for a more fundamental reason.
Suppose that you approach the origin along the curve $y=\sin x$. Then by l’Hospital’s rule you have
$$\begin{align*}
\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{x^3+\sin^3x}{x^2-\sin^2x}&=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{3x^2+3\sin^2x\cos x}{2x-2\sin x\cos x}\\\\
&=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{3x^2+\frac32\sin2x\cos x}{2x-\sin2x}\\\\
&=\lim_{\langle x,y\rangle\to\langle 0,0\rangle}\frac{6x-\frac32\sin2x\sin x+3\cos2x\cos x}{2-2\cos2x}
\end{align*}$$
which does not exist: the numerator approaches $3$ and the denominator, $0$. The problem is that this path, although it stays within the domain of the function, approaches the line $y=x$ so quickly as it approaches the origin that the denominator approaches $0$ much faster than the numerator, and therefore the function blows up as we approach the origin along this path.
