I don't quite understand what we are trying to show when we find an $N$ such that $\forall n\geq N\in\mathbb{R}$ we have $|a_n-L|< \varepsilon$. For example I have been experimenting a bit with the function $$f(n)=\frac{n}{n+1}$$
When put into the definition of the limit $\Rightarrow |\frac{n}{n+1}-1|= |-\frac{1}{n+1}|=\frac{1}{n+1}<\varepsilon$, So now we take $N$ such that $\forall n\geq N$ we have $\frac{1}{n+1}\leq\frac{1}{N+1}<\varepsilon$ which gives us $N>\frac{1}{\varepsilon}-1$, now I'm not entirely sure as to why we don't just use this expression with $n$ instead of $N$, I suppose what I'm asking is why do we bother to introduce this special $N$ when we could just show that for every $\varepsilon$ we can find $n\in\mathbb{N}$ such that $|f_n-L|<\varepsilon$
For example we could just take $\varepsilon>\frac{1}{n+1}>\frac{1}{n+2}$ so then for every $\varepsilon >0$ if we take $n=\frac{1}{\varepsilon}-2$ we clearly have an $n$ that will satisfy any epsilon.
We want to express the idea of "this is true for all n that are significantly large enough". This won't be true for all n, just those that are larger than a certain threshold. That threshold will change depending on the value of $\epsilon$ (in general the threshold will get larger as $\epsilon$ gets smaller). The N is simply the threshold.