Understanding "Equality" notation in a limit I am wondering about the word "equal" in the context of a limit. 
When a limit "equals" a value, is the expression including the $\lim$ equal to the limiting value in the same sense that $1+1 = 2$? 
I am not sure if, in the context of a limit, the limiting value is ever reached. If it is never reached, is the expression with the $\lim$ actually equivalent to the limiting value? 
I am wondering if there are two different types of equality--one in the context of limits and the other in the context of addition. 
Thank you!
 A: I'm not sure what you're asking. But I'll give it a shot (or throw something out there at least!).
Consider the function $f$ defined by
$$
f(x) = 
\begin{cases}
\dfrac{x^2 - 4}{x-2} &\text{if $x \neq 2$,}\\[1ex]
1 &\text{if $x = 2$.}
\end{cases}
$$
The function $f$ is defined for all real numbers; i.e., its domain is the set of all real numbers. Now, we have $f(2) = 1$, yet
$$
\lim_{x \to 2} f(x) = \lim_{x \to 2} \dfrac{x^2 - 4}{x-2} = \lim_{x \to 2} \dfrac{(x+2)(x-2)}{x-2} = \lim_{x \to 2} {} (x+2) = 2 + 2 = \color{blue}{\boldsymbol{4}}.
$$
Thus, we see that $\lim_{x \to 2} f(x)$ is not equal to the value of $f$ at $2$.
If you plot the graph of $f$, you'll actually plot the line given by $y = x+2$, but there will be a hole at the point whose $x$-coordinate is $2$. Is the function still defined at $2$? Yes; $f(2) = 1$.
If we had defined $f$ so that, rather than $f(2) = 1$, we had $f(2) = 4$, then the hole in the graph of $f$ would disappear! That is, the hole in the graph would disappear if we had instead defined $f$ by
$$
f(x) = 
\begin{cases}
\dfrac{x^2 - 4}{x-2} &\text{if $x \neq 2$,}\\[1ex]
\color{blue}{\boldsymbol{4}} &\text{if $x = 2$.}
\end{cases}
$$
Then $f$ would become continuous at $2$. 
What sorts of function (whose domains and ranges are sets of real numbers) have all their values equal to their limits at places? Continuous functions; those are the sorts of functions where you get the same value for the function and for the limit.
A: I will try to give some clarity here based on whatever I make of your question.

First observe the notation of limit $$\lim_{x \to a}f(x) = L\tag{1}$$ Here $f$ is a function of type $f:A \to B$ where $A, B$ are generally subsets of $\mathbb{R}$. Moreover it is important that the domain $A$ should contain an open interval containing point $a$ under consideration except that $A$ need not contain the point $a$ itself. So in order to talk about the limit of $f$ at $a$ (or limit of $f(x)$ as $x \to a$) we must ensure that $f$ is defined in some neighborhood of $a$ except possibly at $a$. The fact that $f$ is defined or not defined at $a$ does not have any significance when we are dealing with limit of $f$ at $a$. Thus even if $f$ is defined at $a$ the value $f(a)$ does not have anything to do with the limit of $f$ at $a$.
Next suppose that the limit $(1)$ exists. Then it means that the whole expression $\lim_{x \to a}f(x)$ is just a real number and we can work with it like real numbers so that I can write $2 + \lim_{x \to a}f(x)$ or $(\lim_{x \to a}f(x))^{2}$. But the expression $\lim_{x \to a}f(x)$ is different from $f(x)$ (that's why we write it in a very different form). Somehow you seem to have the idea that $\lim\dots$ thing can not be used with $=$. This is not correct. The equality sign in equation $(1)$ has the same meaning/usage as in $1 = 2 - 1$ provided that the limit exists. If the limit exists it is a real number. Period.
Another set of problems can be created when we try to treat the expression $\lim_{x \to a}f(x)$ as a real number even when the limit does not exist. And it is best to avoid such mistakes.
The number $L$ on the right of equation $(1)$ is just another real number and equation $(1)$ merely says that the two real numbers $\lim_{x \to a}f(x)$ and $L$ are equal (just like $2 = 4/2 = (\sqrt{2})^{2}$). Some people say this like "$L$ is the limiting value of $f(x)$ as $x \to a$". The meaning of the previous sentence is same as that of equation $(1)$. It does not mean that $L$ is some special number and $f(x)$ can never reach/attain that value and it is only in the limiting sense that $f$ takes the value $L$. Sorry such language is utter non-sense for a person studying calculus.
The fact that $f(x)$ attains (or does not attain) the value $L$ is irrelevant as far as equation $(1)$ is concerned and you need to get used to this if you feel uncomfortable about it.
A: I've had this same question myself. As I understand it, the limit is the real number $L$ that $f(x)$ get's $\mathit{most}$ $\mathit{arbitrarily}$ $\mathit{close}$ to as $x$ approaches $a$, i.e. there does not exist any other real number that $f(x)$ gets closer to. 
The limit is not making a statement about the "true" value of $f(x)$ that is achieved when $x$ "arrives" at  $a$ (which sometimes coincides with $f(a)$, incidentally). The limiting value is the value that is getting pointed to. It need not be "achieved" in order to $\mathit{be}$ the limit.
Let that sink in.
I think this example illustrates my point best:
When we say that $$\lim_{k\to \infty} \sum_{n=0}^k \frac{1}{2^n} = 2$$ we don't mean to say that the partial sums actually ever reach 2, but that the real number on the real number line that the partial sums get closest to is 2. Not 1.9999999999...98, but 2.
Philosophically, we might wonder if such an infinite sum really, truly $\mathit{equals}$ 2, but the use of limits doesn't answer this. If we wanted to investigate that question, we would perhaps drop the limit out front and write $$\sum_{n=0}^\infty \frac{1}{2^n} = \space ?$$
but dealing with a "concrete" infinity like that hasn't been defined, as far as I'm aware.
Curiously, limits were implemented as a way $\mathit{around}$ this philosophical question regarding the infinite. Using limits makes it so calculus doesn't have to use infinitesimals, as it would happen.
So yes, the limit does really, truly $\mathit{equal}$ 2 in this case, but it would seem it's not actually the limit that you're asking about here, but something that creeps into the realm of meta-mathematics and philosophy.
