1549901369-Elements_of_Real_Analysis__Denlinger_

2.2 Algebra of Limits 67

Let no= max{n 1 ,n2}. Then

n 2:: no =? n 2:: n1 and n 2:: n2

c c

::::} lxn - LI < 2 and IYn -Ml < 2

=? l(xn -L) + (Yn--:-M)I < c (Why?)

=? l(xn+Yn)-(L+M)l<c

Therefore, lim n-+oo (xn + Yn) = L + M = n-+oo lim Xn+ nlim -+oo Yn·

( c) Exercise 11.

(d) Let £ > 0. Since {xn} converges, it is bounded; say that 'Vn E N,
lxnl:::; B, where B > 0.
c
Since Xn --t L , 3n1 EN 3 n 2:: n1 =? lxn -LI< 2 (!M! + l) ·

c

Since Yn--+ M, 3 n2 EN 3 n 2:: n2 =? IYn -Ml< 2 B.

Let no= max{n1,n2}. Then

n ;:::: no =? n ;:::: n1 and n 2:: n2

c c

::::} lxn - LI < 2 (!M! + l) and IYn - Ml < 2 B

c c

::::} (!Ml+ l)lxn - LI< 2 and BIYn -Ml< 2

c c

::::} IMllxn - LI < 2 and BIYn -Ml< 2

c c

::::} IMl!xn - LI< 2 and lxnllYn - Ml< 2

c c

::::} IM!lxn - LI+ lxnllYn - Ml< 2 + 2 = E:

::::} IM(xn - L) + Xn(Yn - M)I < c (Why?)