A function $f:\>x\mapsto f(x)$ given by some expression has a "natural" domain of definition $D(f)$: the set of all $x$ in the realm of discourse (${\mathbb R}$ or ${\mathbb C}$, say) for which $f(x)$ can be evaluated without asking questions. In most cases $f$ is continuous throughout $D(f)$, which means that for all $x_0\in D(f)$, when $x$ is sufficiently near $x_0$ then $f(x)$ is very near to $f(x_0)$.
Now some $f$'s may have "exceptional points" where they are not continuous, e.g., the sign-function, which is defined on all of ${\mathbb R}$, but is discontinuous at $0$. Above all, the set $D(f)$ may have "real" or "virtual" boundary points, where $f$ is a priori undefined. But nevertheless we have the feeling that $f$ has a "reasonable" behavior in the neighborhood of such a point. Examples are $x\mapsto{\sin x\over x}$ at $x=0$ (a "real" boundary point of $D(f)$), or $x\mapsto e^{-x}$ when $x\to\infty$ (here $\infty$ is a "virtual" boundary point of $D(f)$).
All in all the concept of "limit" is a tool to handle such "exceptional", or: "limiting", cases. An all-imortant example is of course the following: When $f$ is defined in a neighborhood of $x_0$ we are interested in the function
$$m:\quad x\mapsto{f(x)-f(x_0)\over x-x_0}$$
which has an "exceptional" point at $x_0$. It is impossible to plug in $x:=x_0$ into the definition of $m$.
This brings me to your point 4. which gets to the heart of the matter. I'd rewrite the central sentence as follows: In the definition of the limit of $f(x)$ for $x\to c$ it says that I can make $f(x)$ as close to the value $L$ as I wish, as long as I'm willing to make $x$ sufficiently close to $c$. The idea is: While it is in most cases impossible to put $x:=c$ in the definition of $f$, we want to describe how $f$ behaves when $x$ is very close to $c$.
You then go on to say that "this definition is supposed to be mathematically rigorous, but using these as close and sufficiently close don't look rigorous to me".
The whole $\epsilon$-$\delta$ business serves exactly the purpose to make the colloquial handling of as close and sufficiently close that you are lamenting rigorous.
Life would be simpler if we could define $\lim_{x\to c}f(x)=L$ by the condition $|f(x)-L|\leq |x-c|$, or maybe $|f(x)-L|\leq 100|x-c|$. But four centuries of dealing with limits have taught us that the $\epsilon$-$\delta$ definition of limit, arrived at only around 1870 or so, captures our intuition about them in an optimal way. It takes care as well of the unforeseeable cases when the error $|f(x)-L|$ can be made as small as we we want, but we need an extra effort in the nearness of $x$ to $c$, e.g., $|x-c|<\epsilon^2$ instead of ${\epsilon\over100}$.
Best Answer
When we say that $8$ is the limit of the function $x \mapsto (2x + 2)$ when $x$ approaches 3, what we mean is that the value of $2x + 2$ gets closer and closer to $8$ as we let $x$ move closer and closer to $3$. More specifically, it says that if we pick any sequence $x_1, x_2, x_3, \dots$ of real numbers converging to $3$, the sequence $2x_1 + 2, 2x_2 + 2, 2x_3 + 3, \dots$ converges to $8$. The whole point of taking a limit rather than just evaluating some function at a point is that the number $8$ does not actually need to appear anywhere in the sequence. For example, the limit of the sequence $$ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \dots$$ is 0, even though 0 never appears in this sequence.
Nevertheless, as pointed out in the comments, it frequently happens that we are taking the limit $\lim_{x \to a}f(x)$ of some continuous function $f: \mathbb{R} \to \mathbb{R}$, where $a$ is some fixed real number at which we want to know the limit (so $a = 3$ in your case.) In this case, it follows essentially by definition of continuity that if we choose $x$ to be closer and closer to $a$, then $f(x)$ gets closer and closer to $f(a)$. So in this case, the limit can be computed very easily by simply plugging in $a$ into your function $f$.
In your case, the function $x \mapsto 2x + 2$ is continuous (since if you change $x$ by a small amount $\epsilon$, the value of $2x + 2$ will change by at most $2\epsilon$) and thus $\lim_{x \to 3} 2x + 2 = 2\cdot 3 + 2 = 8$.