3.1. SEQUENCES AND THEIR LIMITS
41
Theorem (3.1.10). Let (an) and (xn) be sequences in R, lim(an) = 0,
and x 2 R. If for some C > 0 and some m 2 N we have
then lim(xn) = x.
|xn
x|  C|an| 8 n
m,
Proof. Let ✏ > 0 be given. Since lim(an) = 0,
⇣✏⌘
⇣✏⌘
✏
9 Ka
2 N 3 8 n Ka
, |an 0| < .
C
C
C
⇢
⇣✏⌘
Let Kx(✏) = max m, Ka
. Then
C
8n
Kx(✏), |xn
x| |{z}
 C|an|
n m
n
✏
<
C
·
= ✏.
|{z}⇣ ⌘
C
✏
Ka
C
Thus lim(xn) = x.
Example.
⇣ 1 + ( 1)n ⌘
(7) lim
= 0.
n
h
⇣
⌘i
1 1
1
Proof. X = 0, 1, 0, , 0, , . . . , 0, , . . . .
2 3
n
1 + ( 1)n
1 + ( 1)n
1+1 2
0 =

=
!0
n
n
n
n
by Example 3. The result follows from Theorem 3.1.10.
⇤
⇤