If we consider a function $f: R \to R$ and $L\in R$.
Then $L$ is said to be the limit of $f(x)$ as $x$ goes to infinity iff:
$\forall \epsilon \in R: [\epsilon \gt 0 \implies \exists m\in R: \forall x\in R: [m\lt x \implies |f(x)-L|\lt \epsilon]]$
And we can say that $L$ is NOT the limit of $f$ as $x$ goes to infinity iff:
$\exists \epsilon_0 \in R: [\epsilon_0 \gt 0 \land \forall m\in R: \exists x_0\in R: [m\lt x_0 \land |f(x_0)-L| \ge \epsilon_0]]$
I am trying to illustrate the negation with an example:
$f: R \to R$ such that $f(x)=\sin (x)$
If I consider $\epsilon_0 =1/2$, then there always exists $x_0 \in(2n\pi +\pi / 6, 2n\pi + 5\pi / 6) \cup (2n\pi +7\pi / 6, 2n\pi + 11\pi / 6) $ such that $\forall m (\gt 0) \in R : [m\lt x_0 \land |f(x_0)| \ge \epsilon_0]$ holds and thus the limit is not $L = 0$ as $x$ goes to infinity.