3 - Math 310 hw 3 solutions Friday 18 Sept 2009 5.2 8.2 d...

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Unformatted text preview: Math 310 - hw 3 solutions Friday, 18 Sept 2009 5.2, 8.2 d, 8.4 5.2 Give the infimum and supremum of each set listed below: (a) S := {x ∈ R : x < 0} Since S = (−∞, 0), we have inf S = −∞ and sup S = 0. (b) S := {x ∈ R : x3 ≤ 8} Since S = (−∞, 2], we have inf S = −∞ and sup S = 2. (c) S := {x2 : x ∈ R} Since S = [0, ∞), we have inf S = 0 and sup S = ∞. (d) S := {x ∈ R : x2 < 8} √ √ √√ Since S = − 2 2, 2 2 , we have inf S = −2 2 and sup S = 2 2. 8.2 (d) Determine the limit of the following sequence and then prove your claim: dn := 2 We will prove that limn→∞ dn = 5 . Let ε > 0 be given and set 2n+4 5n+2 . N := Let n ∈ N and suppose that n > N . Then, n > dn − 16 25ε 16 . 25ε and so 16 25n < ε. Hence, we have 2 2n + 4 2 16 = − = 5 5n + 2 5 25n + 10 since n > 0 the absolute value is unnecessary 16 16 < < ε. 25n + 10 25n 2 The last inequality follows from our choice of N and the assumption that n > N . Hence, dn → 5 . = 8.4 Let (tn ) be a bounded sequence, i.e., there exists M such that |tn | ≤ M for all n, and let (sn ) be a sequence such that lim sn = 0. Prove that lim(sn tn ) = 0. We may assume that M > 0 (for if M = 0, then for all n ∈ N, tn = 0 and sn tn = 0, so there would be nothing to prove). Let ε > 0. Then since lim sn = 0, there is N such that for n ∈ N, n > N implies ε |sn | = |sn − 0| < . M Let n ∈ N such that n > N ; then by our choice of N we have ε M = ε. |sn tn − 0| = |sn ||tn | ≤ |sn |M < M Hence, lim(sn tn ) = 0 as required. ...
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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.

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