Consider the "formal definition" here https://en.wikipedia.org/wiki/Limit_of_a_sequence. I checked some references and this is often precisely the definition in all words and terms used in this article. I claim that this definition is not precise. And I am sure I am wrong. That is why I would like to ask about it. Consider the "illustration" part. This figure makes the concept very clear. It shows that as $\varepsilon$ decreases $N$ should increase. So this means that (i) $N$ must be a function of $\varepsilon$, and (ii) $N$ must be a decreasing function of $\varepsilon$. Why are these not made part of the formal definition by at least writing $N(\varepsilon)$ instead of $N$ and perhaps adding a sentence that $N$ must be a decreasing function of $\varepsilon$? If these are not made explicit, why is it clear from the formal definition that $x_n$ actually converges to $x$ and $n$ increases?
Understanding the definition of the limit of a sequence
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You are right that $N$ depends on $\epsilon$ - and missing this fact is the cause of a lot of confusion among people learning analysis for the first time. Sometimes it is made explicit: "for all $\epsilon > 0$ there exists an $N$ (dependent on $\epsilon$) such that...". But even when phrased as "for all $\epsilon > 0$ there exists an $N$ such that...", $N$ still depends on $\epsilon$: that's how quantifiers ("for all", "there exists") work, even if it's not made explicit.
One thing that should be pointed out is that $N$ is not unique. Indeed if, for a given $\epsilon$, $N = M$ works[1], then $N = M'$ works for any $M' \geq M$. We could define things such that $N$ was the minimum that worked, but this would make both the definition and proofs involving the definition more convoluted, with very little gain in return.
With this in mind, it's somewhat unhelpful to think of $N$ as a function of $\epsilon$, because "function" suggests there's a unique value of $N$ for each $\epsilon$, which isn't true.
On the topic of $N$ increasing as $\epsilon$ decreases: this is not true in the most pedantic sense: if $x_i = 0$ for all $i$, then we could take $N = 10$ when $\epsilon > 1$ and $N = 1$ when $\epsilon \leq 1$, the definition is satisfied but $N$ is not increasing as $\epsilon$ decreases.
However, in a less pedantic sense, it is true that if $N$ works for $\epsilon$ and $\epsilon' > \epsilon$ then $N$ works for $\epsilon'$ too. But this is a consequence of the definition, and so doesn't need to be included in the definition.
To answer your third question, "why is it clear from the formal definition that $x_n$ actually converges to $x$ as $n$ increases?", there's a temptation to ask "what do you mean by 'actually converges'?" Perhaps a helpful observation is that the definition is equivalent to saying: for any $\epsilon > 0$, all but finitely many of the terms of the sequence are within $\epsilon$ of $x$. We don't care how many that "finitely many" is, provided there are only finitely many of them.
[1]: I hope it's clear what I mean by "$N$ works for $\epsilon$" here and later, I just don't want to have to keep writing out the second half of the definition
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The wikipedia article states that we call $x$ the limit of the (real) sequence $(x_n)$ if the following condition holds:
$∀ ε > 0 ( ∃ N ∈ \mathbb N ( ∀ n ∈ \mathbb N ( n ≥ N ⇒ | x_n − x | < ε ) ) )$.
Which is, contrary to your claim, is precise and complete. You are not interpreting the basic logical syntax correctly, which is why you do not understand that the order of quantifiers completely determines the correct interpretation. Here in particular, the "$∃ N ∈ \mathbb N (\cdots)$" is under the "$∀ ε > 0$", so the $N$ that is claimed to exist is not necessarily the same for different $ε$.
Also, there is no logical or pedagogical need to think of $N$ as a function of $ε$. But if you wish, you can do so via Skolemization. However, I recommend you wait until you learn first-order logic before you go into that.
$N$ need not be a function of $\epsilon$. For a given $\epsilon$, many $N$ are possible (in particular all values larger than $N$ also work).
Enforcing $N$ to be some function a $\epsilon$ wouldn't be a good idea, because it would impose a constructive solution, i.e. require you to exhibit the particular value of $N(\epsilon)$. Though in reality, it suffices to prove that a suitable value exists, whatever it is.