{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 4 - Math 310 hw 4 solutions Wednesday 30 Sept 2009 8.8a...

This preview shows page 1. Sign up to view the full content.

Math 310 - hw 4 solutions 8.8a, 8.10, 9.1b; Wednesday, 30 Sept 2009 8.8 (a) Prove the following: lim n →∞ h p n 2 + 1 - n i = 0 . We first observe that p n 2 + 1 - n = p n 2 + 1 - n n 2 + 1 + n n 2 + 1 + n = ( n 2 + 1) - n 2 n 2 + 1 + n = 1 n 2 + 1 + n . Let ε > 0 and set N := 1 ε . Let n N and suppose that n > N . Then we have (note n 2 + 1 + n > 0): p n 2 + 1 - n - 0 = 1 n 2 + 1 + n = 1 n 2 + 1 + n < 1 n < ε. Hence, n 2 + 1 - n 0. 8.10 Let ( s n ) n be a convergent sequence and suppose that lim n →∞ s n > a . Prove that there exists a number N such that n > N implies s n > a . Set s denote the limit of the convergent sequence ( s n ) n . Then by assumption, s > a and so ε := s - a > 0. By definition of limit there exists a number N such that for all n N , n > N implies | s n - s | < ε . Let n N and suppose that n > N , then we have a = s - ε < s n < s + ε. Hence, s n > a as desired. 9.1 (b) Using the limit theorems 9.2-9.6 and 9.7, prove the following. Justify all steps. lim n →∞ 3 n + 7 6 n - 5 = 1 2 . We will need the facts, the first was proved in class (see also Example 9.7(a)) and the second is obvious, lim n →∞ 1 n = 0 lim n →∞ c = c where c is a constant. Observe that our sequence may be rewritten as follows:
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online