Consider:
$$ f(x) = \frac{\sin(\frac{\pi}{x})}{\sin(\frac{\pi}{x})} $$
This function is undefined at $x=0$ and at $ x = \frac{1}{n} $, $\forall n \in \mathbb{N}.$
I thought I could show that the limit of f does not exist at 0, but it does not hold under the appropriate definition of a limit.
Erroneous "proof":
$$ \text{For any arbitrarily small } \delta. $$
$$ M =\bigg \lceil\frac{1}{\delta}\bigg\rceil \in \mathbb{N} $$
$$ \frac{1}{M+1} < \delta $$
$$ f(\frac{1}{M+1}) \text{ is undefined. } $$
$$ \text{Therefore, for any real number L that we choose as our candidate for the limit:} $$
$$ |\frac{1}{M+1}-0| < \delta \text{, but: } $$
$$ |f(\frac{1}{M+1})-L| \nless \epsilon $$
$$ \text{ because it is undefined.} $$
Again, this "proof" is incorrect. According to the definition of the limit for a real-valued function with domain $S\subset\mathbb R$, at a limit point of $S$ (here $S=\{x\in\mathbb R^*\mid\frac1x\notin\mathbb N\}$): $$\text{the limit of our function }f\text{ at }0\text{ is }1\text{, i.e.}$$ $$\forall\varepsilon>0\quad\exists\delta>0\quad\forall x\in S\quad(|x|<\delta\Rightarrow|f(x)-1|<\varepsilon)$$ (indeed, any $\delta>0$ will do, since $\forall x\in S\quad|f(x)-1|=0$).
What I find interesting is that the function becomes "infinitely porous" as we approach 0. It's almost like the line is disintegrating. It reminds me of the function:
$$ g(x) = 1, x \in \mathbb{Q} $$
$$ g(x) = 0, x \not \in \mathbb{Q} $$
I am having trouble wrapping my head around this function. It "behaves" like a normal line everywhere, except infinitesimally close to 0. I want to learn more but I don't know where to begin.
Q1) I'm sure there must be other discussions about functions with these properties, but I cannot find any. Are there any key words I could use to start my research?
Q2) Does this type of discontinuity have a name? I don't think it would be accurate to call it a hole.