Let (xn) be a bounded sequence. Recall that lim inf r, = sup inf xm = sup inf{xn, Xn+1,· ·}, lim sup a, = - inf sup am = infsup{xn, Xn+1,.… · }. m>n n m>n • Show that L= lim infxn is the smallest limit point of (xn), i.e., any subsequential limit of (xn) is not smaller than L, and there is a subsequence that has L as its limit. Remark: Similarly, lim sup xn is the largest limit point of (xn). This also shows that a sequence (xn) converges if n+00 and only if its lim sup and lim inf are equal. Let (xn) and (Yn) be bounded sequences and xn < Yn- Show that lim infxn < lim inf yn - • Let (xn) and (Yn) be bounded ces. Show that lim inf x, + lim inf yn < lim inf(xn + Yn) < lim infxn +lim sup yn, n+00 n00 n00 n+00 n+00 Hint: for parts 2) and 3), you can take some proper subsequences and use part 1).}