Question 1. (a) Give an example to show it is possible to have lim(xn+yn) exist without having limxn or limyn exist.
(b) Give an example to show it is possible to have lim(xnyn) exist without having limxn or limyn exist.
Solution. (a) Consider xn=(−1)n, yn=−xn. Then xn+yn=0 for all n∈N, so lim(xn+yn)=0. But limxn does not exist, because there are two subsequences x2n=1 and x2n−1=−1, which have different limits (1 and -1, respectively). Similarly yn does not have a limit.
(b) Take xn=yn=(−1)n. As was shown above, this sequence does not have a limit. Nevertheless xnyn=(−1)2n=1 for all n, therefore, lim(xnyn)=1.