Actually, you're negating the incorrect definition. You're using the definition for the function $f(x)$ being continuous at some point $p$ of its domain. However, that's not what you're considering. Rather, you're trying to show that there is no value $L$ toward which $f(x)$ tends as $x$ decreases without bound.
If there were such a value $L\in\Bbb R,$ that would mean that we could ensure that $f(x)$ is as close to $L$ as we like, so long as we make $x$ sufficiently negative. Symbolically, $$\exists L\in\Bbb R:\forall\varepsilon>0,\exists N<0:\bigl(x<N \implies|f(x)-L|<\varepsilon\bigr).$$
To prove this false, we must demonstrate the negation $$\forall L\in\Bbb R,\exists\varepsilon>0:\forall N<0,\neg\bigl(x<N \implies|f(x)-L|<\varepsilon\bigr),$$ or equivalently, $$\forall L\in\Bbb R,\exists\varepsilon>0:\forall N<0,\exists x:(x<N)\wedge \neg\bigl(|f(x)-L|<\varepsilon\bigr).$$
So, we let $L\in\Bbb R$ and $N<0$ be arbitrary, and show that there is some $\varepsilon>0$ and some $x<N$ such that $|f(x)-L|\ge\varepsilon.$
It actually turns out to work no matter what $\varepsilon>0$ we choose, fortunately, so we may as well pick an easy one, like $\varepsilon=1.$
Keep in mind that $$|f(x)-L|\ge1$$ is equivalent to $$(f(x)-L\le-1)\vee(f(x)-L\ge1),$$ or $$(f(x)\le L-1)\vee(f(x)\ge L+1).$$ The graph suggests that $f(x)\ge L+1$ is what we want to use, here.
So, given our arbitrary $L\in\Bbb R$ and $N<0,$ can you show that there is some $x<N$ such that $$2+3x^2\ge L+1?$$