Could someone please check my work

For each sequence, find the set S of subsequential limits, the limit superior, and the limit inferior. 1. Let Sn = (- 1)". . The limit superior of S,, is: 0 -1 1 e Do DNE . The limit inferior of Sn is: 0 -1 1 e 0 DNE . Explain: S = { lim (sak), lim (82x- 1} = { - 1, 1} (Sn) = ( - 1)" for all n E N (82k) = (1, 1, ...) = k for all k E N (82k- 1) = ( - 1, - 1, ...) = - k for all k E N The limit superior of (Sn) is 1 because there is a subsequence ($2k) of (Sn) that converges to 1 and no other subsequence of (Sn) converges to higher than 1. In other words, the limit superior of (Sn) is sup(S) = 1 by the definition of limit superior (definition 4.4.9) and supremum of a set, S . Likewise, the limit inferior of (Sn) is -1 because there is a subsequence ($2k - 1) of (Sn) that converges to -1 and no other subsequence of (Sn) converges to lower than -1. In other words, the limit inferior of (Sn) is inf(S) = - 1 by the definition of limit inferior (definition 4.4.9) and infimum of a set, S . 2 . Let (tn ) = o The limit superior of tn is: DNE 0 1/2 1 e co O o The limit inferior of tn is: DNE 0 1/2 1 e co O O OO . Explain: S = { lim (t2k), lim (tzk-1), ...} = {0, 0...} (tn) = = for all n E N (t2k ) = (1, 3, 5,..) = 1 2 k - 1 for all k E N (tzk - 1 ) = 1 1 1 (2' 4' 6'. 2 k for all k E N The limit superior of (tn) is 0 because (tn) converges to 0 and none of the other infinitely many subsequences of (tn) converge to a number higher than 0. In other words, the limit superior of (tn) is sup(S) = 0 by the definition of limit superior (definition 4.4.9) and supremum of a set, S . Likewise, the limit inferior of (tn ) is 0 because (tn ) converges to 0 and none of the other infinitely many subsequences of (tn) converge to a number lower than 0. In other words, the limit inferior of (tn) is inf(S) = 0 by the definition of limit inferior (definition 4.4.9) and infimum of a set, S