Why are "two sided" limits not defined at endpoints? My book gives a non-standard but equivalent definition of a limit:

Let S be a subset of $\mathbb{R}$, let $a$ be a real number or symbol $\pm \infty$ that is the limit of some sequence in $S$, and let $L$ be a real number or symbol $\pm \infty$. We write $\lim_{x \to a^S} f(x) = L$ if $f$ is a function defined on $S$, and for every sequence $(x_n)$ in $S$ with limit $a$, we have $\lim_{n \to \infty} f(x_n) = L.$
Elementary Analysis, 2nd by Kenneth Ross, Page 153

Everybody agrees that if we take $f(x) = x^2$ with domain $\mathbb{R}$, we have $\lim_{x \to 2} f(x) = 4$. But if we restrict the domain of $f$ to $[2, 5]$, I've been told that $\lim_{x \to 2} f(x)$ does not exist. However, I don't see how $\lim_{x \to 2} f(x) = 4$ fails the definition. If $\lim_{x \to 2} f(x)$ does not exist, then there must be some sequence $(x_n)$ on $[2, 5]$ which converges to $2$, but $f(x_n)$ does not converge to $4$. What is this sequence?
I know that such a sequence doesn't exist, but I don't see where my reasoning is wrong.
I could rephrase this question in terms of $\epsilon - \delta$, but I think this definition of a limit is easier to discuss.
 A: There is a disagreement between introductory calculus and real analysis. 
The Calculus definition is: 
"If $a$ lies in some open interval within the domain of $f(x)$, we say that $\lim_{x\to a} f(x)=L$ provided that $f(x)$ gets close to $L$ as $x$ gets close to $a$".
Note that it is phrased in a way for a "first year" student to be able to understand it.
The Analysis definition is: 
"Let $D\subseteq \mathbb{R}$ and $f:D\to \mathbb{R}$. We say, for each $a\in D$ that $\lim_{x\to a}f(x)=L$ if for each $\epsilon>0$, there is some $\delta>0$, such that for Every $x\in D\cap (a-\delta, a+\delta)$, we have $|f(x)-L|<\epsilon$."
If $a$ is at a boundary of the domain, your limit exists according to the analysis definition, but not the Calculus definition. That's why the intro Calculus course should modify the definition in their ciriculum.
A: Basically, the only thing that goes wrong while using the standard definition of limit, is that you need your function to be $defined$ on a two-sided neighborhood of your point (of course if the point is not $\infty$).
So as in your case the function is not defined for $x<2$, the limit cannot exist.
Basically the difference is due to the fact that in your definition you go into the subset topology, whereas in the standard definition, you always work on the full-space topology, so some things fail to make sense.
A: First things first: Under any definition of limit the function $$f:\quad [2,5]\to{\mathbb R},\qquad x\mapsto x^2$$
has $\lim_{x\to2} f(x)=4$.
I don't know about the standing of the passage outlined in skin-tone  in Ross' book. At any rate I'd consider it as an equivalent description of the notion of $\lim_{x\to a}f(x)=L$, but not as definition, whether $a$ is an ordinary number or $a\in\{-\infty,\infty\}$.
Defining limits in terms of sequences is an old habit of old teachers that should be definitely abolished. Defining limits in terms of sequences ultimately means that you would have to test more sequences than there are atoms in the universe in order to prove a single instance of $\lim_{x\to a}f(x)=b$. I'm writing this because you yourself seem to think that limits in terms of sequences are "easier to discuss".
The point is: Limits are not easy to grasp when you are confronted with the exact definition for the first time, because of the nesting of various quantifiers: "For any $\epsilon>0$ there is a $\delta>0$ such that for all $x$ satisfying $\ldots$". Formulating this in terms of sequences adds at least three nested quantifiers to the definition – a nightmare.
Limits are best explained in terms of neighborhoods: "For any neighborhood $V$ of $b$ you can find a neighborhood $U$ of $a$ such that $f\bigl(\dot U\bigr)\subset V\ $". If you can define neighborhoods in terms of a metric you then arrive at the $\epsilon$-$\delta$-definition.  It means that you only have to check whether certain inequalities hold in the given instance.
