Proof of a limit of a function. Is it correct? Edit: There is an answer at the bottom by me explaining what is going on in this post.
Define a function $f : R \to R$ by $f(x) = 1$ if $x = 0$ and $f(x) = 0$ if $x \ne 0$. I was attempting to prove that $\lim_{x \to 0; x\in R}f(x)$ is undefined. The following is my proof.
Proof: Suppose that $\lim_{x\to 0; x \in R}f(x)=L$. Then for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all those $x \in R$ for which $|x-0|<\delta$ we have that $|f(x)-L|<\varepsilon$. But $|0| < \delta$ and by the Archimedean property we know that for $\delta > 0$ there exists an integer $n>0$ such that $0<|\frac 1 n| < \delta$. This is a contradiction, as $f(0) = 1$ and $f(\frac 1 n)= 0$ and both are less than $\delta$.
Is the proof correct?
Edit: Here a limit is defined using adherent points and not limit points. If we were to use limit points then, $\lim_{x \to 0; x\in R\setminus \{0\}}f(x)=0$. I have updated the question with correct notation.
Edit 2: Most textbooks define limits using limit points. In which case you we would have that $\lim_{x \to 0}f(x)=\lim_{x \to 0; x\in R\setminus \{0\}}f(x)$. We are considering the definition of the limit where limits are defined using adherent points. Where it really matters whether we are considering $lim_{x\to 0; x\in R\setminus \{0\}}f(x)$ or $lim_{x\to 0; x\in R}f(x)$.
Edit 3: This is the definition of convergence of a function at a point in the book Analysis 1 by Terence Tao. Let $X$ be a
subset of $R$, let $f : X → R$ be a function, let $E$ be a subset of $X$, $x_0$
be an adherent point of $E$, and let L be a real number. We say that f
converges to $L$ at $x0$ in $E$, and write $\lim_{x \to x_0;x\in E} f(x) = L$, iff $f$, after
restricting to $E$, is ε-close to $L$ near $x_0$ for every $\varepsilon > 0$. If $f$ does not
converge to any number $L$ at $x_0$, we say that $f$ diverges at $x0$, and leave
$\lim_{x\to x_0;x\in E} f(x)$ undefined.
In other words, we have $lim_{x\to x_0;x\in E} f(x) = L$ iff for every $\varepsilon > 0$,
there exists a $\delta > 0$ such that $|f(x) − L| ≤ ε$ for all $x \in E$ such that
$|x − x0| < \delta$.
There are a two other things he defines that are used in this definition. That of $\varepsilon$ closeness and local $\varepsilon$ closeness. For those wanting to read those, Here is the link. It is on page 221.
Edit 4: It might be useless defining limits without limit points. But that is the definition for which I am trying to prove this.
 A: Taking absolute value of positive number os nothing but wasting time!
How you get contradiction from $0<\frac {1} {n} < \delta$ and $f(0) =1$ ?
$1=f(0) <f(\frac{1}{n})=0$ is not true unless $f$ is increasing. In fact $f$ is decreasing on $[0, \infty) $
But choosing $0<\epsilon<1$ , you can produce a contradiction. You have to mention that in your proof as this the most crucial step.
(The way you defined limit)
Let $\epsilon=\frac{1}{2}$ , then $\exists \delta>0$ such that $|x|<\delta \implies  |f(x) -L|<\frac{1}{2}$
For $x=0, |1 -L|<\frac{1}{2}\implies L\in (\frac{1}{2}, \frac{3}{2}) $
For $x=\frac{1}{N}$ (obtained by Archimedean property) , $|L|<\frac{1}{2}$ i.e $L\in (-\frac{1}{2}, \frac{1}{2}) $
Implies $L\in (\frac{1}{2}, \frac{3}{2}) \cap (-\frac{1}{2}, \frac{1}{2}) =\emptyset$ (impossible)
Such $L\in\Bbb{R}$ doesn't exists.
But this is not the correct definition of Limit.
Def:
$\lim_{x\to x_0}f(x) =L$ if $\forall\epsilon>0, \exists \delta>0$ such that $ \forall x\in \Bbb{R} , 0<|x-x_0|<\delta$ implies $|f(x) -L|<\epsilon$
A: 
Proof: Suppose that $\lim_{x\to 0; x \in R}f(x)=L$. Then for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all those $x \in R$ for which $|x-0|<\delta$ we have that $|f(x)-L|<\varepsilon$. But $|0| < \delta$ and by the Archimedean property we know that for $\delta > 0$ there exists an integer $n>0$ such that $0<|\frac 1 n| < \delta$. This is a contradiction, as $f(0) = 1$ and $f(\frac 1 n)= 0$ and both are less than $\delta$.

It's not clear what you are contradicting with. We assume the limit is $L$. Taking $\varepsilon = \frac{1}{5}$, for instance, it should hold that for some $\delta>0$
$$ L-\frac{1}{5} < f(x) < L+\frac{1}{5} $$
for any $x\in (-\delta,\delta)$. In particular, for sufficiently large $N$,
$$ -\frac{1}{5} < 1-L < \frac{1}{5} \quad\mbox{and}\quad -\frac{1}{5} < f(1/N)-L = -L < \frac{1}{5} $$
and that does lead to contradiction. You should write it out.
Furthermore, in your argument, you don't mention what $\varepsilon$ is. It could be $\varepsilon = 100$. I don't see how you arrive at a contradiction.

Alternatively,
Use Proposition 9.3.9. Take the constant sequence $x_n = 0$ for every $n$. Then $f(x_n) \to 1$. For another sequence, $x_n = 1/n$, for instance, $f(x_n) \to 0$. So $\lim _{x\to 0, x\in\mathbb R} f(x)$ does not exist.

This sort of definition for limit is nonstandard. Mainly because it makes the existence of the limit depend on a specific point. It's very dangerous to define it like this. It's not even clear if it is a well defined concept.
A: Tao defines limits using adherent points and not limit points. My previous comments regarding the use of adherent points and limit points throughout this post are completely wrong as I didn’t understand it too well myself at the time either.
But now that I do, here is what is going on in this post. Tao mentions that this is a general definition(a user also commented saying that there are two competing definitions of limits of functions. Unfortunately the answer that comment was under has been deleted. But I think some users with a lot of reputation can still see deleted answers?). He defines a limit as follows(paraphrasing because he used a few simpler concepts to help understand the definition).
Definition: Let $X$ be a subset of $R$. Let $f:X\to R$ be a function. Let $E$ be a subset of $X$. Let $x_0$ be an adherent point of $E$. Let $L$ be a real number. We say that $f$ converges to $L$ near $x_0$ and write $$\lim_{x\to x_0;x\in E}f(x)=L$$ if for every $\varepsilon >0$ there exists a $\delta >0$ such that for all those values of $x\in E$ for which $|x-x_0|<\delta$, we have that $|f(x) - L|<\varepsilon$.
Notice the Specification under the limit sign that $x\in E$. He then remarks that most authors define a limit using limit points. Which in the above notation of a limit would become $$\lim_{x\to x_0;x\in E\setminus \{x_0\}}f(x)=L $$
Notice the specification that $x\in E\setminus \{x_0\}$.
Now to my post, I originally asked if the following limit diverges. $$\lim_{x\to 0;x\in R}f(x) $$ Here we are working with adherent points. In the sense that $x$ can take on the value $0$. Notice the specification that $x\in R$.
This was the first example after defining the limit of a function given by Tao as to what happens to limits if consider $x_0$ to be an adherent point(meaning that we consider the limit $$\lim_{x\to x_0;x\in E}f(x) $$) or a limit point(meaning that we consider the limit $$\lim_{x\to x_0;x\in E\setminus \{x_0\}}f(x) $$). If we consider $0$ above to be a limit point, meaning that we consider the limit $$\lim_{x\to 0;x\in R\setminus \{0\}}f(x)$$ Then that limit converges to $1$. But if we consider $0$ to be an adherent point, meaning that we consider the limit $$\lim_{x\to 0;x\in R}f(x)$$ then the limit is undefined.
As you probably have figured out by now, saying adherent point and limit point when describing whether the value of $x\in E$ or $x\in E\setminus \{x_0\}$ is confusing as limit points are also adherent points. Although, once you get it, it’s clear what it means.
At the time I did not understand this, but then eventually I did. I though I should type up an explanation for the whole confusion I caused that day to others and myself.
