I know what the rules are for negation, so my problem isn't negating -- it's with the interpretation of the negation. We can write $(x_n) \rightarrow x$ as $$\forall \varepsilon > 0 \ \exists n_0 \in \mathbb{N} \ \forall n \ge n_0 \ |x_n-x| < \varepsilon$$ So its negation is: $$\exists \varepsilon > 0 \ \forall n_0 \in \mathbb{N} \ \exists n \ge n_0 \ |x_n-x| \geq \varepsilon$$ In "plain English" I interpret the latter as:
There exists a $\varepsilon$ with the property that for any chosen $n_0$, there is an $n$ such that the difference between $x_n$ and $x$ is greater than $\varepsilon$.
This doesn't make sense to me though -- because $|x_n-x| \geq \varepsilon$ seems to hold for only a single $n > n_0$. If $x$ is not a limit, shouldn't $x_n$ "stay away" from $x$ for all $n > n_0$?
For a number $x$ not to be the limit of $x_n$ we want to have infinitely many terms of the sequence, stay out of some neighborhood of x.
That is exactly what you said in
$$\exists \varepsilon > 0 \ \forall n_0 \in \mathbb{N} \ \exists n \ge n_0 \ |x_n-x| \geq \varepsilon$$
For every $n_0$ there is an $n \ge n_0$ such that , $|x_n-x| \geq \varepsilon$