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Let (xn) be a bounded sequence. Recall that lim inf xn = sup inf xm = sup inf{rn, Xn+1, ...}, lim sup , = inf sup am = inf sup{xn, Cn+1, - .}. +1,•. m>n n m>n n00 n00 n Show that L = lim inf xn is the smallest limit point of (xn), i.e., any subsequential limit of (xn) is not smaller n00 than L, and there is a subsequence that has L as its limit.

Let (xn) be a bounded sequence. Recall that lim inf xn = sup inf xm = sup inf{rn, Xn+1, ...}, lim sup , = inf sup am = inf sup{xn, Cn+1, - .}. +1,•. m>n n m>n n00 n00 n Show that L = lim inf xn is the smallest limit point of (xn), i.e., any subsequential limit of (xn) is not smaller n00 than L, and there is a subsequence that has L as its limit.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let (xn) be a bounded sequence. Recall that lim inf xn = sup inf xm = sup inf{xn, &n+1,..}, lim sup an = inf sup am = infsup{rn, &n+1,• .}. m>n n m>n n00 Show that L = lim inf xn is the smallest limit point of (xn), i.e., any subsequential limit of (xn) is not smaller n-00 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 n00 and only if its lim sup and lim inf are equal. • Let (xn) and (Yn) be bounded sequences and xn < Yn. Show that lim inf xn <lim inf yn . n00 n00 • Let (xn) and (Yn) be bounded sequences. Show that lim inf xn + lim inf yn < lim inf(xn + Yn) < lim inf xn + lim sup Yn, n00 n00 n00 n00 n00 Hint: for parts 2) and 3), you can take some proper subsequences and use part 1).}
Transcribed Image Text:Let (xn) be a bounded sequence. Recall that lim inf xn = sup inf xm = sup inf{xn, &n+1,..}, lim sup an = inf sup am = infsup{rn, &n+1,• .}. m>n n m>n n00 Show that L = lim inf xn is the smallest limit point of (xn), i.e., any subsequential limit of (xn) is not smaller n-00 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 n00 and only if its lim sup and lim inf are equal. • Let (xn) and (Yn) be bounded sequences and xn < Yn. Show that lim inf xn <lim inf yn . n00 n00 • Let (xn) and (Yn) be bounded sequences. Show that lim inf xn + lim inf yn < lim inf(xn + Yn) < lim inf xn + lim sup Yn, n00 n00 n00 n00 n00 Hint: for parts 2) and 3), you can take some proper subsequences and use part 1).}
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