1549901369-Elements_of_Real_Analysis__Denlinger_

2.2 Algebra of Limits 65

Proof. Let { Xn} denote a convergent sequence, say Xn -t L. Taking t: = 1

in Definition 2.1.4, 3 no E N 3 n 2: no ==> lxn - LI < 1. Then,

n ::::-: no ==> -1 < Xn - L < 1 ==> L - 1 < Xn < L + 1."" , ./-/. ( ( d ,_(/t-vh

I ,^1 v-.-J ::...... <-~ • "'.l

(^11) ~ b ~ vv-Jff,•<.A .e ~_6'.:.v t //

. ?" •'-../! \ •'~ ..,._, {/ \C _faa.Ak-C~
Leta=mm{x1,X2,··· ,fu 0 ,L-l}andb=max{x1,··· ,'£rrn,L+l}.Then, )("'"·
'tin EN, a~ Xn ~ b. Therefore, {xn} is bounded. •

Definition 2.2.11 A sequence {xn} is ~unded~y from 0 (by C) if

1 --1• ~ oO--~o_

3C___ >Oand 4 3noEN3n-- 2:no=>lxnl2:C. - 0 cY·1~,, · · - ·

Theorem 2.2.12 (Boundedness Away from 0) If a sequence converges to
a nonzero number, then it is bounded away from 0. More precisely, if Xn -t L =f.
0 and C is any number between 0 and ILi, then {xn} is bounded away from 0
by C. In fact, -1-1 · .... ,..... _,___
(a) If 0 < C < L, then 3no EN 3 n 2: no==> Xn > C.

(. L.

(b) If L < C < 0, then 3 no E N 3 n 2: no ==> Xn < C.

Proof. Suppose Xn -t L =/. 0. Then lxnl -t ILi =/. 0.
Suppose 0 < C < ILi. Then ILi - C > 0. Using Definition 2.1.4, with
t: = ILi - C , 3no EN 3

n 2: no ==> llxnl - ILll < ILi - C

==> c - ILi < lxnl - ILi < ILi - c /

==> C < lxnl < 2ILI - C

==> lxnl > C if.. .;::..

==> Xn > C if 0 < C < L , and Xn < C if L < C < 0.

Therefore, {xn} is bounded away from 0 (by C).^6 •

x,,

I

I

--L

--c

____ J ___ J~------------

____ j ___ J:~----------

Figure 2.3

6. In practice, C is often taken to be ~ or 14-1.