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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?

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    $\begingroup$ No, you can define the limit for any accumulation point of $A$, whether it is in $A$ or not. $\endgroup$ – Thomas Andrews Aug 2 '13 at 11:42
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    $\begingroup$ This is the problem with modern mathematics. We take such a simple problem, and turn it into a technical mess. $\endgroup$ – dezign Aug 2 '13 at 11:44
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    $\begingroup$ @dezign How would you make it simpler? (Hint: Lots of very smart people tried simpler definitions, and they didn't work.) $\endgroup$ – Thomas Andrews Aug 2 '13 at 11:48
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    $\begingroup$ You aapear to mean $A,B\subseteq\mathbb R$, not $A,B\in \mathbb R$. $\endgroup$ – Thomas Andrews Aug 2 '13 at 11:49
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    $\begingroup$ Well, it also says $l\in B$. That is also not part of the definition. It should be $l\in\mathbb{R}$. $\endgroup$ – OR. Aug 2 '13 at 12:37
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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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  • $\begingroup$ I always had the idea from my course of calculus that $\lim_{x\to0}\sqrt{x}$ doesn't exist, so knowing tha actually this limit exists is something really shocking for me and I couldn't believe it when first 18921 told this in his answer and I thought that he was wrong.Now you confirm that this is the case, meaning tha such limit exists, and it's surprising. But also is confusing because for example Siminore have the same idea that such a limit doesn't exist, meaning it doesn't make sense to talk about that. I don't understand. $\endgroup$ – Daniela Diaz Sep 24 '13 at 22:17
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    $\begingroup$ @DanielaDiaz, I think the issue is a schism between modern ideas about limits versus classical ideas. In the classical viewpoint, the order structure of the real line is fundamental to the idea of a limit. Whereas in the modern viewpoint,, its the distance function, or "metric" that really matters. So perhaps the old definitions do imply that the limit fails to exist, however I wouldn't take those definitions too seriously! $\endgroup$ – goblin GONE Sep 24 '13 at 23:27
  • $\begingroup$ @DanielaDiaz I think your calculus teacher was just wrong about this. $\endgroup$ – littleO Sep 25 '13 at 3:01
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I think that yours is essentially a matter of notation. Would it be clearer if $\lim_{x \to 0} \sqrt{x}$ were replaced by $$\lim_{\substack{x \to 0 \\ x \geq 0}} \sqrt{x}?$$ Since we learn limit for functions defined on subsets of $\mathbb{R}$, we tend to think that every independent variable lives in a big set: $\mathbb{R}$. In general topology, the limit is defined via the relative topology of the domain, and there is no confusion.

By the way, a sentence like $\lim_{x \to 0} \sqrt{x}$ is meaningless because we can't consider $x<0$ is just a trap for students: no mathematician would consider it as an important remark! In my opinion, we should make life easier: if a limit is meaningful only from one side, e.g. $x \to a^+$, we could identify $x \to a$ and $x \to a^+$. But notice that $\lim_{x \to -1} \log x$ is meaningless.

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I think the problem is that limits are usually defined only for total functions, rather than partial functions. But, since I've never seen the notion of limit defined for a partial function, I'm going to have to make it up. Here it goes.

Definition. Let $X$ and $Y$ denote metric spaces and $f : X \rightarrow Y$ denote a partial function with domain of definition $A$ such that every element of $X$ is a limit point of $A$. Then, we say that $y \in Y$ is a limit of $f$ at $x \in X$ iff for all $\epsilon > 0$ there exists $\delta > 0$ such that for every $x' \in (B_\delta(x) \setminus \{x\}) \cap A$ we have $f(x') \in B_\epsilon(y).$

Now, I can't promise that this is actually right way of doing things. However, it seems to give the "optimal" answer in a variety of cases. For example, consider the partial function $$f : \mathbb{R} \rightarrow \mathbb{R},\quad f(x \in \mathbb{R} \setminus \{0\}) = \frac{1}{x}.$$

All I mean by this is that $f$ is a partial function, it has domain $\mathbb{R},$ codomain $\mathbb{R}$, and its defined on the set $\mathbb{R} \setminus \{0\}$ and equals $1/x$ there.

Anyway, note that $f(x)$ has no limit as $x$ approaches $0$ according to the definition given, which is what we'd expect.

On the other hand, consider the partial function $$g : [0,\infty) \rightarrow \mathbb{R},\quad g(x \in \mathbb{R}^+) = \sqrt{x}.$$

According to the definition given, $g(x)$ has a unique limit as $x$ approaches $0$, namely $0$. Again, this seems like the "optimal" answer under the circumstances.

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