Trying to understand convergence, example problem with getting an ε

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So I recently started doing convergence in my Algebra class and I am having some trouble understanding conceptually how to find the $\epsilon$-value in the definition $\left|a_n − a\right| \leq \epsilon$.

The script gives the example of

$$\frac{1}{n}\to 0$$

And solves it like this:

$$ \epsilon\lt0,\ N(\epsilon)\in \mathbb{N}$$ with $$N(\epsilon)\geq \frac{1}{\epsilon}.\space\text{Then for}\space n \geq N(\epsilon):$$ $$\left|\frac{1}{n}-0\right|=\frac{1}{n}\leq\frac{1}{N(\epsilon)}\leq \epsilon $$

Where do they take the value of epsilon in relation to N from? Is it already known for this example to just show the concept, or is there some reasoning behind it?

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For a sequence $a_n$, the definition of "$a_n$ converges to $a$" is

For any $\varepsilon>0$, there is an $N\in \Bbb N$ such that for any $n\geq N$ we have $|a_n-a|<\varepsilon$

We want to prove that $\frac 1n$ converges to $0$. To do that, we have to prove that for any $\varepsilon>0$ there is an $N$ such that [...]. One very common way to interpret this is to pretend that we are given a specific $\varepsilon>0$, but we are not told anything else about how large or small it is (it is so-called arbitrary). And then we have to show that there exists an $N$ which works for that $\varepsilon$.

That is what they have done here. Given this $\varepsilon>0$, they pick a corresponding $N$ (they write $N(\varepsilon)$ to emphasise that this $N$ really does correspond to, and is dependent on, $\varepsilon$) to be any natural number larger than $1/\varepsilon$.

So you don't find the $\varepsilon$. It is given to us, and then we find the $N$. On the other hand, if you want to prove that some sequence does not converge to some candidate limit, then it is up to you to show that some $\varepsilon>0$ doesn't work. Usually this means cleverly picking a specific value, tailored to that one problem.