Definition of limit

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.