## University Calculus: Early Transcendentals (3rd Edition)

We $\text{assume}$ that 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 lead us to a contradiction. Thus,the sequence does not have a limit $\{a_{n}\}$ diverges.