Why do limits results change with algebraic manipulation? I just started studying calculus by myself, and I am in limits, but I don't seem to understand why the result changes when I use algebra. I understand how to do it. But why is the result different if it is supposed to be the same expression?
as an example:
$$\lim_{x\to 2} \frac{(x^3-8)}{(x-2)} $$
the limit cannot be solved in this form, but with some algebra
$$\lim_{x\to 2} \frac{(x^3-2^3)}{(x-2)} =\lim_{x\to 2}  \frac{(x-2)(x^2+2x+4)}{(x-2)} = \lim_{x\to 2}  
 (x2+2x+4) = 12$$
I understand that $\frac{a}{0}$ is not determined, and this happens with $\frac{a}{x-2} $ when x=2, and that is why we get rid of that term.
I guess that what we are doing is adding some kind of restriction, where $(x-2)\neq 0$. If that is true, then why we do not specify this in our result? (Or if we are meant to say it, I have not found such reference and I'll be glad if you can refer a source for this to me).
What bothers me is that I have not read any justification about when to use algebraic manipulations further than "to avoid  indeterminates" or "to be able to find the limit" and so on. This helps with its objective, but does not explain why this is valid, or at least I cannot see it.
I am sorry if I am not using the right words. English is my second language.
 A: The limit doesn't change: both of those limits are equal to $12$. It just so happens that substitution doesn't work as a method to calculate the first limit.

I guess that what we are doing is adding some kind of restriction, where $(x−2)\neq 0$. If that is true, then why we do not specify this in our result?

Because limits inherently always have such a restriction: the limit of a function $f$ at a point $x$ is defined as the unique value $y$ such that for all points sufficiently close to $x$ - but not $x$ itself, the value $f(x)$ of the function at that point is close to $y$, if such a thing exists.
This is also the reason that this whole method is valid: when working within a limit as $x \to x_0$, we can assume that $x \neq x_0$, because the limit doesn't care about the value at $x_0$.
For example, if your function is $\mathbb{I}_0$: the function which takes the value $1$ at $0$ and the value $0$ elsewhere, then $$\lim_{x\to 0}\mathbb{I}_0 = 0.$$ Even though your function doesn't take the value $0$ at $0$, the limit does.
Once we've got it into the form of a continuous function (by which I roughly mean "you could draw it without taking your pencil off the paper", with some technical difficulties that mess that up) defined on a neighbourhood of the point (by which I mean you don't have any undefined-ness happening at the point, or at a sequence of points converging to it), you can make use of the fact that the limit of a continuous function at a point that it is defined in a neighbourhood of is equal to its value at that point: since $x^2 + 2x + 4$ is continuous and defined everywhere (unlike $\frac{x^3 - 2^3}{x - 2}$, which is not defined at $x = 2$), we can just substitute in $x = 2$ to find the limit.
If you're interested in going into more detail on this sort of thing and being more careful about these kind of issues, that's generally what is referred to in English as Analysis (as opposed to Calculus, which is the more handwavey type that you've seen so far). There are plenty of online resources for the subject - if you let us know what your first language is, I'm sure someone will be able to suggest resources in that language.
A: You are correct in saying that we are restricting the values of $x$ in the definition of the limit. If we are considering the limit of a function $f$ as $x$ approaches $a$, then we are not actually interested in the value $f(a)$ or even whether $f$ is defined at $a$ at all. In most calculus textbooks the definition of a limit is not actually given but the concept is rather explained intuitively but in the mathematical definition of a limit, this fact is explicitly mentioned. See for example the Wikipedia article on the subject.
The justification for why using algebraic manipulations is allowed can now be better understood. Since we are only concered with values of $f(x)$ for $x\neq a$, we are free to write $f(x)$ in equivalent forms. The limit exists regardless of how we write $f(x)$. It is just that sometimes the limit can be "read off" and other times it will be unclear what the limit is, thus manipulating it algebraically might be necessary.
A: You are under a misunderstanding.
The $\lim_{x \to 2}\frac {x^3-8}{x-2}$ most certainly CAN be solved and when you solve it is $12$.
To solve it:  For any $\epsilon > 0$ no matter how small, if I pick and $x$ so that $2-\sqrt[3]\epsilon < x < 2+\sqrt[3]\epsilon$ and $x \ne 2$ then $12-\epsilon < \frac{x^3-8}{x-2}<12 + \epsilon$.  So that means no matter how close I want to get $\frac{x^3-8}{x-2}$ to $12$ I can do it by forcing $x$ to be close to $2$.  ANd that is what a limit means.
A limit is never about what $\frac {x^3-8}{x-2}$ is when $x$ equals $2$.  A limit is about what $\frac {x^3-8}{x-2}$ is close to when $x$ is close to (but NEVER equal to) $2$.
And as $x$ ISN"T equal to $2$ was can do algebra and "cancel" out that $x-2$ the limit is the same because nothing changed.
