1549901369-Elements_of_Real_Analysis__Denlinger_

68 Chapter 2 11 Sequences

=} JXnYn - LMJ < €.

Therefore, lim (XnYn) = LM = lim Xn· lim Yn·

n-?OO n-+oo n-+oo

(e) Let € > 0. Since Yn --. M -:/=-0, {yn} is bounded away from 0. In fact,
JM[.
by Theorem 2.2.12, 3n1 E N 3 n :'.'.'. n1 =} IYnl > -
2

-. Also, smce Yn --. M,
c[MJ^2.
3n2 EN 3 n :'.'.'. n2 =} IYn - MJ < -
2

-. Let no= max{n1,n2}. Then,

=} IYnl > [~J and IYn - M[ < c[~[

2

1 2 cJM[^2

=} IYn[ < [MJ and IYn - MJ < -2-

[Yn - M[ 1 [Yn - M[ 2 cJMf 2

=} [Yn[[M[ = 1YJ [M[ < [M[. 2[M[

1

1 1 I IM -Yn I IYn - M[

=} Yn - M = YnM = IYn[[M[ < €

=}12__2_1<€.

Yn M

Therefore lim ( 2_) = 2_ =

1

'n->oo Yn M lim Yn.

n->oo

(f) Exercise 14.

(g) Suppose lim Xn = L :'.'.'. 0, and 3 n 1 E N 3 n :'.'.'. n 1 =} Xn :'.'.'. 0. Let € > 0.
n->oo
Case 1 (L = 0): Then 3n2 E N 3 n :'.'.'. n2 =} [xn - Of < c^2. Choose
no= max{n 1 ,n2}. Then

n :'.'.'. no =} 0 :::; Xn < c^2

=} ..;x;;. < €

=} 1..;x;;. - 01 < €.

Therefore, lim n->oo ..;x;;_ = 0 = .JI,.