For almost all great enough natural numbers $n$ we have: $$\frac{n-2}{n^2-4n+2}<0.07$$
I am not sure the "for almost all" here. Does it mean there exists great enough n for which the statement is not satisfied: $$\exists N\in \mathbb{N}: \exists n≥N:\frac{n-2}{n^2-4n+2}≥0.07$$ Or just simply there exists some great enough n for which the statement is satisfied: $$\exists N\in \mathbb{N}: \exists n≥N:\frac{n-2}{n^2-4n+2}<0.07$$ Or something else?
What's meant is
$$\exists N\in \mathbb{N}: \forall n\ge N:\frac{n-2}{n^2-4n+2}<0.07$$
so that the statement holds for all $n$, except at most finitely many (the ones that are $<N$ for which we have no info)
The negation is
$$\forall N\in \mathbb{N}: \exists n\ge N:\frac{n-2}{n^2-4n+2}\ge 0.07$$
which implies there are infinitely many $n$ for which the inequality would fail.
But the first one is true of course and follows from the well-known fact that $$\lim_{n \to \infty} \frac{n-2}{n^2 + 4n +2} = \lim_{n \to \infty}\frac{\frac{1}{n} - \frac{2}{n^2}}{1 + \frac4n + \frac{2}{n^2}} = \frac{0+0}{1+0+0} = 0$$ using that $\frac1n \to 0$ and continuity of arithmetic operations.