Why should this multivariable limit exist? Consider the limit
$$\lim_{(x,y) \to (0,0)} \frac{tan^{-1}(xy)}{xy}$$
My argument for why the the limit does not exist : It does not exist along the path $y=0$.
Or, in another view, $\frac{tan^{-1}(xy)}{xy}$ is undefined on infinite points in any neighborhood of $(0,0)$.
But in many questions like this, the above reasoning is ignored, and we proceed by other techniques.
(Like this : Calculus sin limit with two variables [multivariable-calculus])
But how is that valid ? Can the limit exist with the function undefined in so many points around the given point ?
 A: Here's a definition for limits:

Let $X,Y$ be metric spaces, $E\subseteq X$, $f:X\to Y$ be a function, and $a$ be a limit point of $E$. We say the function $f$ has a limit at $a$ (in the space $Y$) if the following condition is satisfied:

*

*There exists $l\in Y$ such that for every $\epsilon>0$, there exists a $\delta>0$ such that for all $x\in E$, if $0 <d_X(x,a)< \delta$ then $d_Y(f(x), l) < \epsilon$.

In this case, we can prove $l$ is unique and we write $\lim\limits_{x\to a}f(x) = l$

In this formulation of limits, note that the function $f$ does not have to be defined on the whole space $X$. It only needs to be defined on a certain subset $E$ (it is very well possible that $X\setminus E$ is an infinite set, but this doesn't matter). Moreover the point, $a$, where we're calculating the limit doesn't even have to be an element of $E$; we only need $a$ to be a limit point of $E$.
In your case, we take $X=\Bbb{R}^2, Y= \Bbb{R}$ (both with the usual Euclidean metrics) and $E = \{(x,y)\in\Bbb{R}^2| \, xy \neq 0\}$. In this case, we define $f:E\to Y= \Bbb{R}$ by $f(x,y) = \frac{\arctan(xy)}{xy}$, and the point $(0,0)$ is certainly a limit point of the set $E$. Thus, we can certainly try to calculate the limit (and in this case the limit does exist and equals $1$... if you need more elaboration on that let me know)
