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

This preview shows page 21 - 24 out of 85 pages.

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.
Image of page 21
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).
Image of page 22
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 .
Image of page 23
Image of page 24

You've reached the end of your free preview.

Want to read all 85 pages?

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

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes