I think my explanation in class was sub-par at best. In the solution for #1, we distribute a limsup and a liminf over a sum, which is not in general ok to do. The reason it is ok here is because one of the limits exists, as is pointed out in the #1 solution.
$ Lemma : $ Suppose $ a_n \to a $ then $ \liminf (a_n+b_n) = a + \liminf b_n $
$ Proof: $ Given $ \epsilon >0, \exists N $ such that $ \forall n>N, a_n\in B_{\epsilon}(a) $ and $ \inf_{k>n} b_k \in B_{\epsilon}(\liminf_m b_m) $. Hence, $ \liminf (a_n+b_n) = \lim_n \inf_{k>n} (a_k + b_k)\in B_{2\epsilon}(a + \liminf_m b_m) $, and the result follows as $ \epsilon $ is arbitrary.
