Neighborhood of $x$, for $\lim_{x \to a} \sqrt{x}$ I know that for $\lim\limits_{x \to a} \sqrt{x} = \sqrt{a}$.
From the definition of the limit, it implies that if $\lim\limits_{x \to a} f(x)=L\ \text{then} \ \forall \varepsilon \gt 0 \ \exists \delta \gt 0 |\  0\lt |x-a|\lt\delta \implies |f(x)-L|\lt \varepsilon$.
If I take $a=2$, $\lim\limits_{x \to 2} \sqrt{x} = \sqrt{2}$, this would mean I could take a neighborhood of at the max only 2 for x because if I take a neighborhood of more than 2 from the left the neighborhood would enter negative $x$-axis.
So there is a strict restriction on the $\delta$ that I'm considering which would imply a further caveat on values of $\varepsilon$ which implies the neighborhood here does not hold $\forall \varepsilon$.
So there should be an additional statement missing in the definition, or I'm missing something in the interpretation.
 A: This is a fuller definition of $$\lim\limits_{x \to a} f(x)=L,$$ where the domain $D$ of $f$ is a superset of a punctured neighbourhood҂ of $a:$ $$\forall\varepsilon{>}0\;\exists\delta{>}0\;\color{red}{\forall x{\in} D}\,\bigg(0<|x-a|<\delta\implies|f(x)-L|<\varepsilon\bigg).$$

  ҂ For $\lim\limits_{x \to a} f(x)$ to exist, $f$ must be defined on a punctured neighbourhood of $a$—though not necessarily at $a$ itself.
A: There is something missing in the definition, yes. Let $D_f$ be the domian of $f$. Then the definition of $\lim_{x\to a}f(x)=L$ is$$(\forall\varepsilon>0)(\exists\delta>0):0<|x-a|<\delta\color{red}{\wedge x\in D_f}\implies\bigl|f(x)-L\bigr|<\varepsilon.$$
A: I think that you might perhaps be misjudging your quantifiers. When you pick an arbitrary $\epsilon$, you then simply have to show that there exist a single $\delta$ satisfying your definition. In your example you seem to be choosing your $\delta$ first and claiming that an $\epsilon$ does not exist for that value. This is erroneous.
You are correct in that there are restrictions on $\delta$ but that doesn't matter since you simply have to pick a single value once $\epsilon$ has been fixed.
