TITLE: What is $\lim_{(x,y)\to(0,0)} \frac{(x^3+y^3)}{(x^2-y^2)}$?
QUESTION [5 upvotes]: 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!

REPLY [7 votes]: Brian's answer is nice, so it got me thinking, what about approaching the point along other functions, $y=f(x)$, with $f(0)=0$. Let us plug that into the expression to get $\frac{x^3+f^3}{x^2-f^2}$. This will give $$\lim_{x \to 0} \frac{x^2-xf+f^2}{x-f}\;.$$
We will now perform a l’Hospital’s rule to get $$\lim_{x \to 0} \frac{2x-xf'-f+2ff'}{1-f'}\;.$$
If at this point we assume that $f'(0)=1$, we have another $0/0$. If $f'\neq 1$, we get that the limit is zero.
So letus now hit this with another l’Hospital. This will give us
$$\lim_{x \to 0} \frac{2-2f'-xf''+2f''f'+2(f')^2}{-f''}\;. \qquad (*)$$
When evaluate at $x=0, f'=1$, we get  $$\lim_{x \to 0} -\frac{2f''+2}{f''}\;.$$
And you can decide what you want the limit to be by picking a value for $f''$.