Calculus with Analytic Geometry by edwards & Penney soln ch11

Calculus with Analytic Geometry

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Section 11.2 C11S02.001: The most obvious pattern is that a n = n 2 for n 1. C11S02.002: The most obvious pattern is that a n = 5 n 3 for n 1. C11S02.003: The most obvious pattern is that a n = 1 3 n for n 1. C11S02.004: The most obvious pattern is that a n = ( 1) n 1 2 n 1 = 1 2 n 1 for n 1. C11S02.005: The most obvious pattern is that a n = 1 3 n 1 for n 1. C11S02.006: The most obvious pattern is that a n = 1 n 2 + 1 for n 1. C11S02.007: Perhaps the most obvious pattern is that a n = 1 + ( 1) n for n 1. C11S02.008: One pattern is that a n = 15 2 5 2 · ( 1) n . Another is that a n = 5 · 1 + 1 ( 1) n 2 for n 1 . C11S02.009: lim n →∞ 2 n 5 n 3 = lim n →∞ 2 5 3 n = 2 5 0 = 2 5 . C11S02.010: lim n →∞ 1 n 2 2 + 3 n 2 = lim n →∞ 1 n 2 1 2 n 2 + 3 = 0 1 0 + 3 = 1 3 . C11S02.011: lim n →∞ n 2 n + 7 2 n 3 + n 2 = lim n →∞ 1 n 1 n 2 + 7 n 3 2 + 1 n = 0 + 0 + 0 2 + 0 = 0. C11S02.012: This sequence diverges because a n = n 3 10 n 2 + 1 > n 3 10 n 2 + 10 n 2 = n 20 + as n + . C11S02.013: Example 9 tells us that if | r | < 1, then r n 0 as n + . Take r = 9 10 to deduce that lim n →∞ 1 + ( 9 10 ) n = 1 + 0 = 1 . C11S02.014: Example 9 tells us that if | r | < 1, then r n 0 as n + . Take r = 1 2 to deduce that lim n →∞ 2 ( 1 2 ) n = 2 0 = 2 . 1
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C11S02.015: Given: a n = 1 + ( 1) n for n 1. If n is odd then a n = 1 + ( 1) = 0; if n is even then a n = 1 + 1 = 2. Therefore the sequence { a n } diverges. To prove this, we appeal to the definition of limit of a sequence given in Section 11.2. Suppose that { a n } converges to the number L . Let = 1 2 and suppose that N is a positive integer. Case 1: L 1. Then choose n N such that n is odd. Then a n = 0, so | a n L | = | 0 L | = L 1 > . Case 2: L < 1. Then choose n N such that n is even. Then a n = 2, so | a n L | = | 2 L | = 2 L > 1 > . No matter what the value of L , it cannot be made to fit the definition of the limit of the sequence { a n } . Therefore the sequence { a n } = { 1 + ( 1) n } has no limit. (We can’t even say that it approaches + or −∞ ; it does not.) C11S02.016: Because 1 + ( 1) n = 0 if n is odd and 2 if n is even, 0 1 + ( 1) n n 2 n for all n 1. Therefore, by the squeeze law for sequences, lim n →∞ 1 + ( 1) n n = 0. C11S02.017: We use l’Hˆopital’s rule for sequences (Eq. (9) of Section 11.2): lim n →∞ a n = lim n →∞ 1 + ( 1) n n ( 3 2 ) n = lim x →∞ 1 ± x 1 / 2 ( 3 2 ) x = ± lim x →∞ 1 2 x 1 / 2 ( 3 2 ) x ln ( 3 2 ) = 0 . C11S02.018: We use the squeeze law for sequences (Theorem 3 of Section 11.2): 1 sin n 1 for all integers n 0, and therefore 1 3 n sin n 3 n 1 3 n for all integers n 1. By the result in Example 9 of Section 11.2, 1 / 3 n 0 as n + . Therefore lim n →∞ sin n 3 n = 0 . C11S02.019: First we need a lemma. Lemma: If r > 0, then lim n →∞ 1 n r = 0. Proof: Suppose that r > 0. Given > 0, let N = 1 + [[1 / 1 /r ]]. Then N is a positive integer, and if n > N , then n > 1 / 1 /r , so that n r > 1 / . Therefore 1 n r 0 < . Thus, by definition, lim n →∞ 1 n r = 0. 2
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Next, we use the squeeze law for sequences (Theorem 3 of Section 11.2): 1 sin n 1 for all integers n 1, and therefore 0 sin 2 n n 1 n for all integers n 1. But by the preceding lemma, 1 / n 0 as n + . Therefore lim n →∞ sin 2 n n = 0 .
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