State each of the following English sentences in symbols, as its final form, not containing $\lnot$.
(a): $f$ is not continuous at $a$
(b): The sequence $(a_n)_{n \in \mathbb{N}}$ approaches to $\infty$ as $n$ increases.
(c): $\lnot$(b)
Since i know that the formal definition of $f$ being continuos at $a$ is
$\forall \epsilon (\epsilon>0 \rightarrow \exists \delta (\delta>0 \land \forall x (|x-a|< \delta \rightarrow |f(x)-f(a)|<\epsilon)))$
My try is negating the whole statement by: $$\exists\epsilon(\epsilon>0 \land \forall\delta(\delta>0 \rightarrow \exists x(|x-a|<\delta \land |f(x)-f(a)| ))\geq\epsilon)$$ I'm not confident that this is equivalent to $f$ being discontinuous, because I can't come up with a graphical intuition it.
Likewise, if (b) is $$\forall M (M>0 \rightarrow \exists N_0 (N_0 \in \mathbb N \land (\forall n ((n \in N \land n>n_0)\rightarrow f(x)>M ))))$$
is $$\lnot(b) = \exists M(M>0 \land \forall N_0(N_0 \in N \rightarrow (\exists n ((n \in \mathbb N \land n>N_0)\land f(x)\le M))))$$ ?
Both of your solutions are correct, though the second would be better stated without the unnecessary restriction on $M$, as indicated in my comment.
Concerning the "graphical intuition", the definition of continuity can be rephrased as "for every neighborhood of $f(a)$, there is a neighborhood of $a$ which is carried by $f$ into the neighborhood of $f(a)$. In other words: if you draw a circle about $f(a)$, then you can also draw a circle about $a$ such that everything inside it is carried into the circle about $f(a)$.
Your negation says there is some neigbhorhood of $f(a)$ such that any neighborhood of $a$ will contain at least one point that is carried outside the neighborhood of $f(a)$. This is what we mean by discontinuity: no matter how close you get to $a$, something is carried away from $f(a)$.