We first start with the usual definition of limit.
Let $c$ be a real number / point. A neighborhood of $c$ is an open interval $I$ containing $c$. If $I$ is a neighborhood of $c$ then the set $I\setminus \{c\} $ is said to be a deleted neighborhood of $c$.
Let $A$ be a non empty subset of $\mathbb {R} $ and let $f:A\to\mathbb {R} $ be a function. Further let $c$ be a point such that there exists some deleted neighborhood of $c$ contained in $A$.
A number $L$ is said to be the limit of $f$ at $c$ (written symbolically as $L=\lim\limits_{x\to c} f(x) $) if the following condition holds: for every $\epsilon>0$ there exist a $\delta>0$ corresponding to it such that $|f(x) - L|<\epsilon$ whenever $x\in A$ and $0<|x-c|<\delta$.
Such a number $L$ may or may not exist, but if it exists then the above definition ensures that it must be unique (see proof later).
The above definition can be generalized a bit to include domains which are not necessarily neighborhoods. And the question here provides such a general definition using concept of limit points.
The same generalization however does not apply to one sided limits because of lack of corresponding concepts like left / right limit points. The idea of one sided limits is restricted to introductory courses of calculus where typical domains are intervals/neighborhoods. In that context here is a definition of left hand limit.
Let $A$ be a non-empty subset of $\mathbb {R} $ and let $f:A\to\mathbb {R} $ be a function. Further let $c$ be a point such that there exists a number $h>0$ with $(c-h, c) \subseteq A$.
A number $L$ is said to be the left hand limit of $f$ at $c$ (written symbolically as $L=\lim\limits_{x\to c^{-}} f(x) $) if the following condition holds: for every $\epsilon>0$ there exists a $\delta>0$ corresponding to it such that $|f(x) - L|<\epsilon $ whenever $x\in A$ and $0<c-x<\delta$.
The key prerequisite in the notion of limit is that if $c$ is the point under consideration then the function must be defined at points arbitrarily near to $c$. For one sided limits the function must be defined at points arbitrarily near to and on the corresponding side of $c$.
The uniqueness of a limit is based on the fact that values of the function are near the limit $L$. If $L_1,L_2$ are distinct then you can not ensure values $f(x) $ to be arbitrarily near to both $L_1,L_2$ at the same time. If you go too close to $L_1$ then you have to necessarily move away from $L_2$.
This is more technically expressed as follows:
Theorem (with easy proof) : Let $a, b$ be two real numbers with $a\neq b$. Then there exists a neighborhood $I$ of $a$ and a neighborhood $J$ of $b$ such that $I\cap J=\emptyset$.
This key fact is the basis of the general notion of a Hausdorff space.
It seems like you're quoting this passage from Wikipedia. Note the condition stated at the beginning of that paragraph, that $c$ has to be a limit point of $D$. If it's not (like $c=3$ in your example), then it's not meaningful to even begin talking about “$\lim\limits_{x \to c} f(x)$”, so the limit doesn't exist in that case.
By the way, your function is actually continuous (since, well, it satisfies the definition of continuity, as can be verified). Its domain is disconnected, but that's a different concept.
Best Answer
Where you wrote $|x-1|<\delta,$ you need $0<|x-1|<\delta.$
I.e. you can make $f(x)$ as close to the limit as you want by making $x$ close enough, but not equal, to $1.$