Unformatted text preview: n ≥ N . Hence x n = n is both frequently even and odd. This does not conﬂict with 3.2.4 as the ﬁrst case says via 3.2.4 that it is not true that x n = n is ultimately odd, while the second case says that it is not true that x n = n is ultimately even. Page 36 Problem 2. a. Note x n = 1 for n = 4 k + 1 with k ∈ Z . Hence given N take n = 4 N + 1 to get n ≥ N with x n = 1. b. Note ﬁrst that sin x ≤ x for all x ≥ 0 (this is used in Calculus 1 to show that (sin x ) = cos x ). Now 2 π n < 1 for all n ≥ 7, so also sin( 2 π n ) < 1 for all n ≥ 7. 1...
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 Fall '10
 Girardi

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