TITLE: Definition of limit
QUESTION [6 upvotes]: I learnt at school that this limit $\lim_{x\to 0}\frac{1}{x}$ doesn't exist, and intiuitively it seems that such is the case, but I just don't get it. 
To begin with, I understand the definition of limit in this way, please tell me where I'm wrong or if I'm missing something: 

Let $A, B\subseteq \mathbb{R}$ and $f:A\longrightarrow B$ a function such that $a\in A$ is an acummulation point. Then we say that $l\in B$ is the limit of the function $f$ when $x$ approches $a$ and is denoted by $\lim_{x\to a}f=l$ if and only if $\forall \epsilon\in \mathbb{R}(\epsilon>0)\exists\delta\in \mathbb{R}(\delta >0)\forall x\in A(0<|x-a|<\delta\longrightarrow |f(x)-l|<\epsilon)$.

So, accordingly, I have the function $f:\mathbb{R}\setminus\{0\}\longrightarrow\mathbb{R}$ such that $f(x)=\frac{1}{x}$. Since $0\notin Dom (f)$ then it doesn't even make sense to talk about the definition of $\lim_{x\to 0}\frac{1}{x}$.
Also I think that I probably need to change in my definition the part of $(\forall x\in A)$ for $(\forall x\in \mathbb{R})$. This is consistent because if my metric spaces were not subsets of $\mathbb{R}$, for example if I had $E_{1}, E_{2}$ metric spaces and $A\subseteq E_{1}, B\subseteq E_{2}$ such that $f:A\longrightarrow B$. For the part $|x-a|<\delta$ to make sense it's necessary that $x\in A$ or $x\in E_{1}$. The problem here is that taking $(\forall x\in E_{1})$ might turn undefined many points of the part $|f(x)-l|$ because it might be that $A\subseteq E_{1}$ but $A\neq E_{1}$. 
Edit: With all the suggestions - thank you so much guys - my new definition is this way:

Let $A, B\subseteq \mathbb{R}$ and $f:A\longrightarrow B$ a function such that $a\in \mathbb{R}$ is an acummulation point of $A$. Then we say that $l\in \mathbb{R}$ is the limit of the function $f$ when $x$ approches $a$ and is denoted by $\lim_{x\to a}f=l$ if and only if $\forall \epsilon\in \mathbb{R}(\epsilon>0)\exists\delta\in \mathbb{R}(\delta >0)\forall x\in A(0<|x-a|<\delta\longrightarrow |f(x)-l|<\epsilon)$.

Now, I have this problem. With this definition I can prove that given the function $f:\mathbb{R^{+}}\longrightarrow \mathbb{R}$ such that $f(x)=\sqrt{x}$, then $\lim_{x\to 0}\sqrt{x}=0$. But officially this limit doesn't exist, though $\lim_{x\to 0^{+}}\sqrt{x}=0$. 
If I substitute $\forall x\in A$ for $\forall x\in \mathbb{R}$ then the problem seems to be fixed. But now this doesn't allow to talk about rational functions, like for example if I take the function $f:\mathbb{Q}\longrightarrow \mathbb{R}$ such that $f(x)=x$ then $\lim_{x\to 0}f(x)$ doesn't exist. What am I missing?

REPLY [1 votes]: The definition given in your edit is correct.  It agrees with definition 4.1 in baby Rudin.  This definition does imply that $\lim_{x \to 0} \sqrt{x} = 0$, which is a true statement.
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