There are functions which are discontinuous everywhere and there are functions which are not differentiable anywhere, but are there functions with domain $\mathbb{R}$ (or "most" of it) whose limit does not exist at every point? For example, $ f:\mathbb{R}\to\mathbb{N}, f(x) = $ {last digit of the decimal representation of $x$}. Is this even a valid function?
[Math] Function whose limit does not exist at all points
calculuslimitsreal numbers
Related Solutions
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.
Suppose $f$ is differentiable in some neighborhood $(x-\delta,x+\delta)$ of $x$, and $\lim\limits_{t\rightarrow x}{f'(t)}$ exists. Define $y:(x-\delta,x+\delta)\rightarrow\mathbb{R}$ such that $y(t)$ is strictly between $x$ and $t$ and $$f(t)-f(x) = f'(y(t))(t-x)$$ for every $t\in(x-\delta,x+\delta)$. The existence of such a function is guaranteed by the Mean Value Theorem. Since $y(t)$ is between $x$ and $t$ for every $t$, this implies that $\lim\limits_{t\rightarrow x}{y(t)} = x$, and since $y(t)\ne x$ for $t\ne x$ as well, we have $\lim\limits_{t\rightarrow x}{f'(y(t))}=\lim\limits_{s\rightarrow x}{f'(s)}$ by the composition law (think of $s = y(t)$ in this substitution). This implies that $$f'(x) = \lim\limits_{t\rightarrow x}{\frac{f(t)-f(x)}{t-x}} = \lim\limits_{t\rightarrow x}{f'(y(t))} = \lim\limits_{t\rightarrow x}{f'(t)},$$ i.e. $f'$ is continuous at $x$.
Remark: Typically, the composition law is phrased as follows: if $\lim\limits_{x\rightarrow c}{g(x)} = a$ and $f$ is continuous at $a$, then $\lim\limits_{x\rightarrow c}{f(g(x))} = \lim\limits_{u\rightarrow a}{f(u)}$. In our problem, we obviously cannot assume $f'$ is continuous at $x$, since that is what we are trying to show. However, the above conclusion still holds if we merely require that $g(x)\ne a$ if $x\ne c$ in some neighborhood of $c$. A proof of this can be found here (look for "Hypothesis 2").
Best Answer
As suggested in the comments define $f:\mathbb{R}\rightarrow \mathbb{R}$ by $f(x)=0$ if $x$ is irrational and $f(x)=1$ if $x$ is rational. Let's prove it doesn't have a limit at any point. Let $y\in \mathbb{R}$, suppose $y$ is irrational and that there exists $L$ the limit of $f$ at $y$. Then given any $\epsilon<1/2$, there is $\delta>0$, such that $x\in (y-\delta, y+\delta)\setminus\{y\}$ implies $|f(x)-L|<1/2$. Now, since there exists $x_0,x_1\in (y-\delta, y+\delta)\setminus\{y\}$ such that $x_0$ is rational and $x_1$ is irrational we get that $1=|1-0|=|f(x_0)-f(x_1)|\leq |f(x_0)-L|+|f(x_1)-L|<1/2+1/2=1$, a contradiction. So the limit at any irrational $y$ does not exists. The same argument applies to any rational $y$, so the limit doesn't exists at every point.