Can a limit exist if there are paths to the limit point where the function is not defined? For example, does the limit of $f(x,y) = \frac{bxy}{xy}$ for any constant $b$ exist for $(x,y) \to (0,0)$?
Does the fact that for $x=0$ and $y=0$ you have a problem with deviding by zero imply that there is no limit?
Or could you extend the definition of a limit with "for all $x$ and $y$ in the domain of $f$..."?
In fact, this question arose from the following question: 

For which constants $a$, $b$ and $c$ does the limit $(x,y) \to (0,0)$ exist for $f(x,y) = \frac{ax^2+bxy+cy^2}{xy}$?

When approaching $(0,0)$ through lines $y=mx$, it turns out that the limit depends on $m$ if not both $a$ and $c$ are $0$. So the first condition for the existing of a limit is that both $a$ and $c$ are $0$. 
That leaves the question for which $b$ the limit $(x,y) \to (0,0)$ exist for $f(x,y) = \frac{bxy}{xy}$. And this is where we started doubting the solution. We found 3 approaches, which we don't know is the right one. Can you simply say: 


*

*Hey, $f(x,y) = \frac{bxy}{xy}$ simply results to $b$, so the limit exists for every $b$ (and equals $b$)? Or should you say:

*$f(x,y) = b$ for $x \ne 0$ and $y \ne 0$, and $f(x,y)$ is undefined for $x = 0$ or $y = 0$. And this makes the limit nonexistent, because for pairs $(x,y)$, even though close to $(0,0)$, you cannot guarantee anything about $|f(x,y)-b|$ because if either $x = 0$ or $y = 0$ the function isn't even defined properly.  Or is it the case that:

*$f(x,y) = b$ for $x \ne 0$ and $y \ne 0$, and $f(x,y)$ is undefined for $x = 0$ or $y = 0$. And this is no problem, the limit still exists for every $b$ (and is $b$)?
 A: Let's take the metric space point of view. $\mathbb{R}^2$ has the standard (Euclidean) metric (i.e. distance function): $d((x,y),(a,b)) = \sqrt{(x-a)^2+(y-b)^2}$ and $\mathbb{R}$ has the standard metric $d(x,a)=\sqrt{(x-a)^2}=|x-a|$.
Let $A \subseteq \mathbb{R}^2$. Then $A$ is itself becomes a metric space when we restrict the  metric to $A$ (i.e. restrict the domain of $d$ from $\mathbb{R}^2 \times \mathbb{R}^2$ to $A \times A$. 
Now given a function $f:A \to \mathbb{R}$, $\lim\limits_{(x,y) \to (a,b)} f(x,y)=L$ if and only if for each $\epsilon >0$ there exists a $\delta >0$ such that for each $(x,y) \in A$ such that $0<d((x,y),(a,b))=\sqrt{(x-a)^2+(y-b)^2}<\delta$ we have $d(f(x,y),L)=|f(x,y)-L|<\epsilon$.
So for your example, the domain of $f$ is clearly $A=\{(x,y)\;|\; x\not=0 \mathrm{\;and\;} y\not=0 \}$. And as you mention, if $x\not=0$ and $y\not=0$, then $f(x,y)=\frac{bxy}{xy}=b$. Thus $\lim\limits_{(x,y)\to (0,0)} f(x,y) = \lim\limits_{(x,y)\to (0,0)} b =b$.
Thus the limit exists when working within the subspace $A$. Although there still is something a little fishy here since we are limiting to $(0,0)$ which does not belong to $A$!
Now if you are working in $\mathbb{R}^2$. Then the function isn't even defined on any open ball (with deleted center) centered at the origin, so there is no $\delta>0$ such that for all $0<d((x,y),(0,0))<\delta$ we have that $f(x,y)$ is even defined. So in this sense, the limit does not exist. 
However, we can easily "repair" the definition of $f$, extending it to all of $\mathbb{R}^2$ using the definition:
$$f(x,y) = \left\{ \begin{array}{cc} \frac{bxy}{xy} & x \not=0 \mathrm{\;and\;} y \not=0 \\ b & x=0 \mathrm{\;or\;} y=0 \end{array} \right. $$
which of course is equivalent to $f(x,y)=b$ (everywhere). Then the limit exists as before.
So in the end which answer is correct? Exists? Doesn't? 
Well, it depends on your definitions. Usually calculus texts are fairly sloppy when it comes to these matters. It's anyone's guess as to whether the text would say it exists or not. My guess is if you pick a random book you've got a 50/50 chance of either being told "No. The limit doesn't exist because the function isn't defined on a deleted neighborhood of the origin." OR "Yes. The limit exists because $f(x,y)=b$ (after carelessly canceling off '$xy$')."
As for me, I would tend to interpret such a problem in the "sloppy sense" where we "repair" $f$ and treat it as the constant function $f$. If I were in a picky mood, I would object to the initial question since strictly speaking $f$ is not a function defined on a domain which includes the limit point.
A: As André Nicolas suggests in the comments, a pragmatic approach is "to take a more generous point of view, and to quietly remove removable singularities."
To put this on a rigorous footing, we need a bit of topology.
Let $S$ and $C$ be topological spaces, and let $f: D \to C$ be a function defined on a subset $D \subset S$ and taking values in $C$.
Let $L \subset S$ be the set of limit points of $D$.  We may define the limit of $f$ at a point $a \in L$ as a value $y \in C$ such that, for every neighborhood $V \subset C$ of $y$, there exists a neighborhood $U \subset S$ of $a$ such that, for all $x \in U \cap D \setminus \lbrace a \rbrace$, $f(x) \in V$.
This definition naturally generalizes the notion of a limit of a function to (almost) arbitrary topological spaces.  The only requirement we need to impose, in order to ensure that the limits thus defined are unique, is that $C$ should be a Hausdorff space.
In your example above, $S = \mathbb R^2$, $C = \mathbb R$ and the function $f$ is defined on $D = (\mathbb R \setminus \lbrace 0 \rbrace)^2$, so that the set $L$ of limit points of $D$ is all of $\mathbb R^2$.
We may immediately note that, for all points $(x,y) \in D$, the function $f(x,y) = \frac{bxy}{xy} = b$ is constant.  Thus, $f$ has a limit at every point in $L = \mathbb R^2$, and the limit equals $b$.
