Answer
$\{a_{n}\}$ diverges.
Work Step by Step
ASSUMPTION:
The sequence $ \{a_{n}\}=\{0,2,0,2,0,2,...\}$ has a limit L.
Then, by definition, if we choose $\epsilon=1$, there must exist an index N such that
for $n\gt N$ (for sequence terms after the Nth), $|L-a_{n}|\lt 1.$
After the Nth term, there are infinitely many terms that equal 0.
It must be then, that $|L-0|\lt 1.$
That is, $L\in(-1,1).$
After the Nth term, there are infinitely many terms that equal $2$.
It must be then, that $|L-2|\lt 1.$
That is, $L\in(1,3).$
But, L can't be an element of both these intervals, since their intersection is empty.
Thus, our assumption has shown a contradiction, and it can't be true.
The sequence does nos have a limit.
$\{a_{n}\}$ diverges.
