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

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Unformatted text preview: Math 310 - hw 4 solutions Wednesday, 30 Sept 2009 8.8a, 8.10, 9.1b; 8.8 (a) Prove the following: n→∞ lim n2 + 1 − n = 0. We first observe that √ n2 + 1 + n (n2 + 1) − n2 1 n2 + 1 − n √ =√ =√ . n2 + 1 + n n2 + 1 + n n2 + 1 + n √ Let ε > 0 and set N := 1 . Let n ∈ N and suppose that n > N . Then we have (note n2 + 1 + n > 0): ε n2 + 1 − n = n2 + 1 − n − 0 = Hence, √ n2 + 1 − n → 0. √ n2 1 1 1 =√ < < ε. 2+1+n n +1+n n 8.10 Let (sn )n be a convergent sequence and suppose that limn→∞ sn > a. Prove that there exists a number N such that n > N implies sn > a. Set s denote the limit of the convergent sequence (sn )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 |sn − s| < ε. Let n ∈ N and suppose that n > N , then we have a = s − ε < sn < s + ε. Hence, sn > a as desired. 9.1 (b) Using the limit theorems 9.2-9.6 and 9.7, prove the following. Justify all steps. 1 3n + 7 =. lim n→∞ 6n − 5 2 We will need the facts, the first was proved in class (see also Example 9.7(a)) and the second is obvious, 1 lim =0 lim c = c n→∞ n n→∞ where c is a constant. Observe that our sequence may be rewritten as follows: 3n + 7 3n + 7 = · 6n − 5 6n − 5 1 n 1 n = 1 3 + 7n sn 1=t 6 − 5n n 1 n 1 1 where sn := 3 + 7 n and tn := 6 − 5 n . By Theorem 9.2 and the fact that → 0, we have 1 1 = 7 lim =7·0=0 n→∞ n n 1 1 lim −5 = −5 lim = −5 · 0 = 0. n→∞ n→∞ n n Hence, by Theorem 9.3 and the above fact concerning the limit of a constant we have 1 1 lim sn = lim 3 + 7 = lim 3 + lim 7 = 3 + 0 = 3, n→∞ n n→∞ n→∞ n n→∞ 1 1 lim tn = lim 6 + (−5) = lim 6 + lim −5 = 6 + 0 = 6. n→∞ n→∞ n→∞ n n→∞ n 1 Finally, since 6 = 0 and for all n ∈ N, tn = 6 − 5 n = 0, we may apply Theorem 9.6 to conclude that n→∞ lim 7 3n + 7 sn limn→∞ sn 3 1 = lim = ==, n→∞ tn 6n − 5 limn→∞ tn 6 2 where the Theorem is applied at the equation marked ( ). n→∞ lim ...
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This note was uploaded on 03/26/2010 for the course MATHEMARIC 131a taught by Professor Janeday during the Spring '10 term at San Jose State University .

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