Iterated and Double limits exisitence

Question

The existence of the double limit of a multi variable function at a point implies that the iterated limits will also be the same in value and exist, given that we know that that the individual single limits of the function, in x and y exist. $$\lim_{(x,y) \to (\alpha, \beta)} f(x,y) = L\implies \lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) = \lim_{y \to \beta} (\lim_{x \to \alpha} f(x,y)) = L$$ Given that we know that : $$\lim_{x \to \alpha} f(x,y) \mbox{ and } \lim_{x \to \beta} f(x,y) \mbox{ exist } $$ My question is why is this so? Why is the existence of single limits necessary for the above mentioned implication.
As an example take the following function: $$ f(x,y) = xsin(1/y) $$ $$\lim_{(x,y) \to (0,0)} f(x,y) = 0 $$ $$\therefore \mbox{ as expected :} $$ $$ lim_{y \to 0} (\lim_{x \to 0} f(x,y)) = 0 $$ $$ \mbox{but} $$ $$ lim_{x \to 0} (\lim_{y \to 0} f(x,y)) = \mbox{Does not exist} $$

Answer 1

One of the iterated paths doesn't exist because the line $y = 0$ is not in the domain of $f(x,y)$. So, $\lim_{{y}\to{0}}f(x,y)$ doesn't exist. But, we can still travel through every other path within the domain of $f(x,y)$ and the limit remains $0$.

You're not meant to follow paths outside the domain to check if a limit exists.