Can I evaluate a multivariable limit by using paths instead of factoring? I had a question in my calculus exam which was:
Evaluate the limit
$$\lim_{(x,y)\to(0,0)} \frac{x^2-y^2}{3x-3y}$$
My solution was by using (along $y$ and along $x$) paths and I got $0$ in both paths.
However, I got marks deducted because my calculus doctor said that I must factor first
instead of using along $y$ and along $x$ paths. So is my solution mathematically incorrect? Am I required to do factoring, or is it okay to use paths?
 A: You are fine to use paths, in theory, but if you take that approach you would need to show that the limit is zero for every path to the origin.  Simply showing that you have two paths which both give you the same limit isn't enough.  So, for this one, since you are trying to show the limit exists it would probably be simpler to show it directly (by factoring e.g.).  Using paths is frequently a good method for showing that a limit does not exist, because in that case you only need to find two paths with different limits.
A: You should exclude points $y=x$ i.e. $(x,x)$, because function is not defined there. For other points you have
$$\lim_{(x,y)\to(0,0)} \frac{x^2-y^2}{3x-3y}= \lim_{(x,y)\to(0,0)}\frac{x+y}{3}=\lim_{(x,y)\to(0,0)}\frac{x}{3}+\lim_{(x,y)\to(0,0)}\frac{y}{3}=0$$
So factoring here, imho, is best way.
A: You may consider the limit along two paths converging to $(0,0)$ to show the limit does not exist. However, you can't use this to show the limit exists.
I would like to address another issue here. To say
$$ \lim_{(x,y) \to (0,0)} f(x,y) = \ell $$
is to say that $\ell-\epsilon<f(x,y)<\ell+\epsilon$ for any $\epsilon>0$, for all points $(x,y)$ within some ball centered at $(0,0)$, disregarding the center $(0,0)$. The radius of the ball would vary with $\epsilon$, and often with the center point $(0,0)$ as well.
Now every ball centered at $(0,0)$ must contain points on the line $y=x$. You can see this geometrically, and write a simple equation to show this algebraically. Thus, $f(x,y)$ is undefined at some point $(x,x)$ within every ball centered at $(0,0)$. So there is no question of the limit existing.
