Proof Let M 0 Select a real number m so that 0 m lim t n Whether lim t n or not

# Proof let m 0 select a real number m so that 0 m lim

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Proof Let M > 0. Select a real number m so that 0 < m < lim t n . Whether lim t n +∞ or not, it is clear that there exists N 1 such that n > N 1 implies t n > m ; see Exercise 8.10. Since lim s n +∞ , there exists N 2 so that n > N 2 implies s n > M m . Put N max { N 1 , N 2 } . Then n > N implies s n t n > M m · m M . Example 7 Use Theorem 9.9 to prove that lim n 2 + 3 n + 1 +∞ ; see Example 6. Solution We observe that n 2 + 3 n + 1 n + 3 n 1 + 1 n s n t n where s n n + 3 n and t n 1 1 + 1 n . It is easy to show that lim s n +∞ and lim t n 1. So by Theorem 9.9, we have lim s n t n +∞ . Here is another useful theorem. 9.10 Theorem. For a sequence ( s n ) of positive real numbers, we have lim s n +∞ if and only if lim( 1 s n ) 0 . Proof Let ( s n ) be a sequence of positive real numbers. We have to show lim s n +∞ implies lim 1 s n 0 (1) and lim 1 s n 0 implies lim s n +∞ . (2) In this case the proofs will appear very similar, but the thought processes will be quite different.
2. Sequences 52 To prove (1), suppose that lim s n +∞ . Let > 0 and let M 1 . Since lim s n +∞ , there exists N such that n > N implies s n > M 1 . Therefore n > N implies > 1 s n > 0, so n > N implies 1 s n 0 < . That is, lim( 1 s n ) 0. This proves (1). To prove (2), we abandon the notation of the last paragraph and begin anew. Suppose that lim( 1 s n ) 0. Let M > 0 and let 1 M . Then > 0, so there exists N such that n > N implies | 1 s n 0 | < 1 M . Since s n > 0, we can write n > N implies 0 < 1 s n < 1 M and hence n > N implies M < s n . That is, lim s n +∞ and (2) holds. Exercises 9.1. Using the limit theorems 9.2–9.6 and 9.7, prove the following. Justify all steps. (a) lim n + 1 n 1 (b) lim 3 n + 7 6 n 5 1 2 (c) lim 17 n 5 + 73 n 4 18 n 2 + 3 23 n 5 + 13 n 3 17 23 9.2. Suppose that lim x n 3, lim y n 7 and that all y n are nonzero. Determine the following limits: (a) lim( x n + y n ) (b) lim 3 y n x n y 2 n 9.3. Suppose that lim a n a , lim b n b , and that s n a 3 n + 4 a n b 2 n + 1 . Prove lim s n a 3 + 4 a b 2 + 1 carefully, using the limit theorems. 9.4. Let s 1 1 and for n 1 let s n + 1 s n + 1. (a) List the first four terms of ( s n ). (b) It turns out that ( s n ) converges. Assume this fact and prove that the limit is 1 2 (1 + 5).
Exercises 53 9.5. Let t 1 1 and t n + 1 t 2 n + 2 2 t n for n 1. Assume that ( t n ) converges and find the limit. 9.6. Let x 1 1 and x n + 1 3 x 2 n for n 1. (a) Show that if a lim x n , then a 1 3 or a 0. (b) Does lim x n exist? Explain. (c) Discuss the apparent contradiction between parts (a) and (b). 9.7. Complete the proof of 9.7(c), i.e., give the standard argument needed to show that lim s n 0. 9.8. Give the following when they exist. Otherwise assert “NOT EXIST.” (a) lim n 3 (b) lim( n 3 ) (c) lim( n ) n (d) lim(1 . 01) n (e) lim n n 9.9. Suppose that there exists N 0 such that s n t n for all n > N 0 . (a) Prove that if lim s n +∞ , then lim t n +∞ . (b) Prove that if lim t n −∞ , then lim s n −∞ . (c) Prove that if lim s n and lim t n exist, then lim s n lim t n .

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• Spring '14
• Limits, Limit of a sequence, Sn, lim sn

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