If ∃ denotes "there exists" and ∀ denotes "for all", then your version of negation of the limit definition is

∃ε ∀n_{0}∀n. n > n_{0}⇒ |x_{n}- 1/2| ≥ ε

where x_{n}= (3n-2)/(4n+1). I wrote ∀n because you did not provide a quantifier for n, and by default the quantifier is universal. In fact, the correct negation is

∃ε ∀n_{0}∃n. n > n_{0}and |x_{n}- 1/2| ≥ ε.

In words, there exists an ε such that there exist arbitrarily far elements of the sequence whose distance to 1/2 is at least ε. One way to prove that there exist arbitrarily far elements of the sequence with some property P is to show thatallelements possess this property starting from some point. That is, to prove

∀n_{0}∃n. n > n_{0}and P(x_{n}) (1)

we show

∃n_{0}∀n. n > n_{0}⇒ P(x_{n}) (2)

because (2) is stronger that (1). (Why?)

To prove (2), note that x_{n}approaches 3/4 from below. Therefore, eventually it is greater than say, the midpoint between 1/2 and 3/4, i.e., 5/8. So, take ε = 5/8 - 1/2 = 1/8 and find an n_{0}such that x_{n}≥ 5/8 for all n > n_{0}.