Why do we "plug" the approaching value when solving limits? This is a question regarding basic limits.
If we have
$\lim_{x\to3} (2x + 2)$
then,
$\lim_{x\to3} (2x + 2) = 2(3) + 2 = 8$
Now, we know that x approaches 3, but never actually gets to 3 (correct?). So, if x never gets to 3, then why do we use the approaching value itself when solving the limit, when it's just close to 3, but not equal?
 A: 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 $\lim_{x \to 3} 2x + 2 = 2\cdot 3 + 2 = 8$.
