134 Chapter 2 • Sequences
Therefore, since limits preserve inequalities, lim Xnm ~ lim Xm and
m-+oo m-+oo
lim Xnm ::::; lim Xm. Putting these inequalities together with (a), we have
m-+oo m--+cx:>
the desired inequalities, lim Xn ::::; lim Xmk ::::; lim Xnk ::::; lim Xn· •
n--+ oo k--+ 00 k--+ oo n--+ ex:>
Theorem 2.9.7 (.s Criterion for Upper Limit) Let {xn} be a bounded
sequence. Then L = lim Xn ¢=> Ve > 0,
n-+oo
(a) Xn < L + e, for all but finitely many n, and
(b) Xn > L - e for infinitely many n.
Proof. Part 1 ( =?): Suppose L = lim Xn and let e > 0. By definition,
n-+oo
this means L = lim Xn. Then
n-+oo
(a) By definition of limit, :J no E N 3
n ~ no =? lxn - LI < e
=? L - e < sup{xk: k ~ n} < L + e.
=? L + e is an upper bound for {xk: k ~ n}.
Then, '</k ~ n 0 , Xk < L+e. That is, Xn < L+e for all but finitely many n.
(b) By Remark 2.9.5 (b), L = inf{xn : n E N}. Let no E N. Then
Xn 0 > L - e/2, and since Xn 0 = sup{xk: k ~no}, :J k ~no 3 Xk > Xn 0 - e/2.
Thus, '</no EN, :J k ~no 3 Xk > L -e. That is, Xn > L - e for infinitely many
n.
Part 2 (~): Suppose that Ve> 0, the given conditions (a) and (b) hold.
Let e > 0. Then by (a), :J no EN 3
n ~ no =? Xn < L + e
=} Xn::::; L + e.Thus, Ve> 0, lim Xn ::::; L + e, so by the forcing principle, lim Xn ::::; L.
n--+ oo n--+ ex>
On the other hand, by (b), '<In EN, Xn = sup{xk : k ~ n} ~ L - .s. Thus,
since limits preserve inequalities, lim Xn ~ L, so lim Xn ~ L. •
n--+ ex> n--+ oo
Theorem 2.9.8 (.s Criterion for Lower Limit) Let {xn} be a bounded
sequence. Then L = lim Xn ¢=> Ve > 0,
n-+oo
(a) Xn > L - .s for all but finitely many n, and
(b) Xn < L + e for infinitely many n.
Proof. Exercise 3. •