Prove $\lim_{x\to 0}$ of piecewise $f(x)=1$ if $x=1/n$ where $n\in \mathbb{Z}$ or $0$ otherwise on $[-1,1]$ doesn't exist. $$f(x)=\begin{cases} 1 & \text{for } x=\frac{1}{n}, \; n\in\mathbb{Z}\\0 & \text{otherwise.}\end{cases}$$
I want to prove with the Epsilon-Delta definition  that the limit as $x \to 0$  doesn't exist so I can say there is an essential discontinuity at $x=0$.
My reasoning is:
$\lim_{x\to 0}  f(x) \not= L$
Given any $\epsilon > 0$ , we can take  $\delta >0$ for which:
There exists $0 < |x| < \delta$ $\to$ $|f(x)-L|$ $≥$ $\epsilon$
So I'm trying to say that there are values of $x$ within $(-\delta, \delta)$ for which  $|f(x)-L|$ $≥$ $\epsilon$. I think what I wrote might be saying that, for any $\delta$ i choose, all values within $(-\delta, \delta)$ will result in  $|f(x)-L|$ $≥$ $\epsilon$, which I think isn't true no matter how infinitely small our chosen $\delta$ is.
How do I properly prove this?
 A: To prove that the limit doesn't exist, you shouldn't start with Given any $\epsilon \gt 0$... but by finding an epsilon for which the definition of the limit won't be satisfied.
So suppose that the limit exists and its value is $L$. Now take $\epsilon = 1/3$. We have to prove that whatever $\delta \gt 0$ we pick, it exists $0 \lt \vert x \vert \lt \delta$ such that $\vert f(x) - L \vert \ge \epsilon = 1/3$.
Now it is quite simple and you can separate two cases.
Case 1: $L \lt 1/2$
Then you can find $n \in \mathbb N$ such that $0 \lt 1/n \lt \delta$ and you have
$$\vert f(1/n) -L \vert = \vert 1 -L \vert \gt 1/2 \gt 1/3.$$
Case 2: $L \ge 1/2$
In that case pick $n \in \mathbb N$ such that $0 \lt \sqrt 2/n \lt \delta$. Now
$$\vert f(\sqrt2/n) -L \vert = \vert 0 -L \vert \ge 1/2 \gt 1/3.$$
Note: I picked up $\epsilon = 1/3$ because in all neighborhood of $0$ $f$ takes values at distance equal to $1$. And so, whatever $L$ would be, some function values would be at more than $1/3$ distance from the potential limit.
