Are Conditional Statements Automatically Universals?

Question

I am trying to negate the following statement: "For every positive number $\epsilon$, there is a positive number $M$ for which $|f(x)-b|<\epsilon$ whenever $x>M.$" I write this in symbols as: $$(0) \ \forall \epsilon>0 \exists M>0 (x>M \implies |f(x)-b|<\epsilon). $$ Now it is unclear to me whether the appropriate negation of this statement is: $$(1) \ \exists \epsilon>0 (\forall M>0 \exists x (x>M \wedge |f(x)-b|\geq \epsilon))$$ or $$(2) \ \exists \epsilon>0 (\forall M>0 (x>M \wedge |f(x)-b| \geq \epsilon). $$ It feels natural to introduce the $\exists x$, but formally speaking, the first statement did not include a $\forall x$ quantifier, so I'm not sure whether it's justified to introduce a $\exists x$ in the negation. What is the correct negation of $(0)$?

Answer 1

$(2)$ is the negation of $(0)$.

The negation of a formula has the same occurences of bound and free variables.

In $(0)$ and $(2)$ variable $x$ is free, but in $(1)$ it is bound.