Answers & Hints for Selected Exercises 641(1) Using algebra and Thm. 2.2.13,1. 4n^2 - 5n 4 - .2.
im = lim n
n->= 8n^2 + 3n - 1 n-><Xl 8 + l n - ~ n
lim (4 - .2.)
n--+oo n
n-+oo lim (8 + l n - ~) n
lim 4 - lim .2.
n-+oo n-+oo n 4 -^5 n-+oo lim 1. n^1
lim 8 + lim l - lim ~
2 - -.
8 + 3 lim 1. - ( lim 1.)
2
n--+oo n--+oo n n-+oo n n--+oo n n-+oo n- In each case, prove the sequence is unbounded. For example, in ( c),
2n+3 Vn > 11:?:. Vn = (^2) yi• /n.
- By Thm. 2.2.16, only the tail of {xn} matters.
- Apply the algebra of limits to Exercise 1.2-B.6.
EXERCISE SET 2.3- Apply the first squeeze theorem with an = 0, bn = Ian - LI and en = lbn I·
Then apply Thm. 2.2.l (b). - (a) I 2:7n - ol = 2:1n < 7~ < ~ -t 0.
(c) I 3n"+s - ~1=I3n;~~8+B) I= 3n8+s < 3~ = ~ · ~ -t O.
(e) I lOn-11 7-2n _ (-s) I = I (10n-11)+5(7-2n) 7-2n I = 2n__1!_ -7 < --1L 2n-n =^24 n ...... 0(g) I~ -o1 = ~ ...... o.
(for n > 7).
(i)^0 < n2+5 n < 1i2" n -- n 1. -t^0 ·
(k) I 3n(^2) +n-5 31 = I (3n^2 +n-5)-3(n^2 +6n) I = 1-17n-5 I = 1 7n+ 5 < 17n+n
n^2 +6n n2+6n n2+6n n2+6n ~
= ~ -t 0. (m) I 2n~~7 1 = 2n
5
3~7 < 2nf~n3 = ~ -t 0.
(for n ~ 2)
(o) I n
(^3) -2n (^2) .!
1
12(n^3 - 2n^2 )-(6-3n+2n^3 ) I - I -4n^2 +3n-6
1
1 4n
2
- 3n+6 1
6 -3n+ 2n3 2 - 2(6-3n+2n3) - 2(2n3-3n+6) - 2 l(2n3-3n+6) 1
= ln(4n-3)+61 = 4n^2 - 3n+6 < 4n^2 = 4n^2 = 2 -t 0
2 ln(2n2-3)+6 1 2(2n3-3n+6) 2(2n3-3n2) 2n2(2n- 3) 2n-3 ·
(for n ~ 2)
- Modify the proof of (a) where appropriate.
- (b) By Exercise 1.3.4, and the algebra of limits,
r 12 +2^2 +3^2 + .. +n^2 - r (n(n+l)(2n+l) 1 ) - r (n n+l 2n+l 1)
n.:.1! n - n.:.1! 6. 71:3 - n.:.1! n. n-. n. 6
= 1. 1. 2. i = ~-
en - (a) Let p EN be fixed and Vn EN, Xn =-.Note that when lei= 1,
nP