The ε–N definition of a limit

Definition. $u_n\to\ell$ means: for every $\varepsilon>0$ there exists $N$ such that $n\ge N\Rightarrow|u_n-\ell|<\varepsilon$.

1. We have $$\left|\dfrac{3n+1}{n+2}-3\right|=\dfrac{5}{n+2}<\dfrac5n, \, \, \forall n \geq 0.$$
Given $\varepsilon>0$ take $N=\lceil5/\varepsilon\rceil+1$; for $n\ge N$ we get $\frac5n<\varepsilon$, hence
$\left|\dfrac{3n+1}{n+2}-3\right| \leq \varepsilon$.

2. Suppose $(-1)^n\to\ell$. With $\varepsilon=1$ there is $N$ such that $|(-1)^n-\ell|<1$ for $n\ge N$. Then for $n\ge N$,
$$2=\left|(-1)^{n+1}-(-1)^n\right|\le\left|(-1)^{n+1}-\ell\right|+\left|\ell-(-1)^n\right|<1+1=2,$$
which is absurd. So the sequence diverges.

Remark. In part 2 the contradiction arises because the terms stay at distance $2$ from each other forever, whereas convergent sequences have terms that eventually cluster.