Clarification about $\forall x$ in the definition of limit The limit for $x\to x_0$ where $x_0 \in \mathbb{R}$ of $f(x)$ resulting in a finite limit, that is to say $\ell \in\mathbb{R}$ is defined as:

we write $\lim_{x\to x_0} f(x) = \ell$ where $\ell \in\mathbb{R}$, if $\forall \epsilon > 0,\ \exists \delta_{\epsilon} >0$ such that $\forall x \in\mathbb{\Omega}$ there exists a neighbourhood of $x_0$, that is $0 < |x-x_0| < \delta_{\epsilon}$ for which we have $\vert f(x)-\ell \vert < \epsilon$

Where $\Omega$ is the domain of $f(x)$.
In what sense we need to have $\forall x \in\Omega$?
For example: let's verify the truthness of the limit
$$\lim_{x\to 1} 1+x = 2$$
Then
$$\vert 1+x - 2 \vert < \epsilon$$
That is
$$-\epsilon +1 < x < \epsilon +1$$
Which as $\epsilon \to 0$ this is indeed a neighbourhood of $1$.
Yet I do not understand why $\forall x$.
 A: As José Carlos Santos pointed out, the "there exists a neighborhood" part of the definition of limit is poorly worded to the point of being arguably wrong. With minimal changes, the definition should read something like:

We write ${\displaystyle \lim_{x\to x_0}}f(x)=\ell$ where $\ell\in\mathbb R$, if $\forall \epsilon>0$, $\exists \delta_{\epsilon}>0$ such that $\forall x\in\Omega$: if $x$ lies in the "neighbourhood" of $x_0$ that is (given by) $0 < |x-x_0| < \delta_{\epsilon}$ then we have $\vert f(x)-\ell \vert < \epsilon$.

To answer the question "Why $\forall x\in\Omega$?" there are (at least) three things that need to be addressed:

*

*Why can't we just leave it without a quantifier on $x$?

*Why isn't the quantifier $\exists$ instead of $\forall$?

*Why does the $\Omega$ matter?

To answer #1, it's because every variable in mathematics needs to be defined or quantified, and (to simplify the logic a bit) implicit quantifiers are $\forall$ that come before everything else. So if we don't put a quantifier for $x$, then we would implicitly assume that to capture the idea of limit, we'd need the entirety of "$\forall \varepsilon>0,\exists \delta_{\varepsilon}>0$ such that if $0<|x-x_0|<\delta_{\varepsilon}$ then $|f(x)-\ell|<\varepsilon$" to hold for each $x$. For example, if $x=7$ and $x_0=3$, then $0<|x-x_0|<\delta_{\varepsilon}$ is satisfied for $\delta_{\varepsilon}=5$, say, but $|f(7)-\ell|$ might not be less than $\varepsilon$ if $f$ is almost any function, even one that should really have a limit at $x_0=3$.
To answer #2, it's because $\exists$ would just mean that there's an $x$ near $x_0$ where $f$ takes a value near $\ell$. But if $f$ varies wildly, maybe there are lots of values of $x$ near $x_0$ where $f$ doesn't. For example, if $f(x)=\left(\sin(1/x)\right)/x$, then using $\exists$ would let us say that the limit is $17$ since there are values (e.g. approx. $.0503$, $.0475$, $.0387$, $.0313$...) close to $x_0=0$ where $f(x)=17$. But that would not match our intuitive idea of limit since the function also attains extremely large and extremely small values near $x_0=0$.
To answer #3, it helps to look at a function whose domain is not most of the real line. Consider $f(x)=\sqrt{x}+3$ for $x\ge0$. If we just wrote $\forall x$ instead of $\forall x\in\Omega$ (implicitly thinking $\forall x\in\mathbb R$), then to examine the limit at $x_0=0$, we'd have to consider negative $x$ in $0<|x-0|<\delta_{\varepsilon}$. But for those negative $x$, we'd never have $|f(x)-\ell|<\varepsilon$ since $|f(x)-\ell|$ wouldn't even be defined for negative $x$. This basically means that nice functions with restricted domains like $\sqrt{x}+3$ couldn't have limits at the edge of their domain, when we'd intuitively want ${\displaystyle \lim_{x\to 0}}\sqrt{x}+3$ to be $3$. For this reason, we restrict to only caring about $x\in\Omega$.
