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Final 01

# Final 01 - Su Yung Final 1 Due 11:00 pm Inst Shinko Harper...

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Su, Yung – Final 1 – Due: Dec 14 2007, 11:00 pm – Inst: Shinko Harper 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the value of lim x 0 e 2 x - e - 2 x sin 4 x . 1. limit = 2 3 2. limit = 1 correct 3. limit does not exist 4. limit = 4 5 5. limit = 3 2 6. limit = 5 4 Explanation: Set f ( x ) = e 2 x - e - 2 x , g ( x ) = sin 4 x. Then f, g are everywhere differentiable func- tions such that lim x 0 f ( x ) = lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule applies: lim x 0 f ( x ) g ( x ) = lim x 0 f 0 ( x ) g 0 ( x ) . Now f 0 ( x ) = 2( e 2 x + e - 2 x ) , g 0 ( x ) = 4 cos 4 x, while lim x 0 f 0 ( x ) = 4 , lim x 0 g 0 ( x ) = 4 . Consequently, lim x 0 e 2 x - e - 2 x sin 4 x = 1 . keywords: 002 (part 1 of 1) 10 points Determine whether the sequence { a n } con- verges or diverges when a n = ( - 1) n 3 n + 4 7 n + 3 , and if it does, find its limit. 1. limit = 3 7 2. limit = 0 3. limit = 4 3 4. sequence diverges correct 5. limit = ± 3 7 Explanation: After division, 3 n + 4 7 n + 3 = 3 + 4 n 7 + 3 n . Now 4 n , 3 n 0 as n → ∞ , so lim n → ∞ 3 n + 4 7 n + 3 = 3 7 6 = 0 . Thus as n → ∞ , the values of a n oscillate be- tween values ever closer to ± 3 7 . Consequently, the sequence diverges . keywords: 003 (part 1 of 1) 10 points

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Su, Yung – Final 1 – Due: Dec 14 2007, 11:00 pm – Inst: Shinko Harper 2 To apply the root test to an infinite series n a n the value of ρ = lim n → ∞ | a n | 1 /n has to be determined. Compute the value of ρ for the series X n = 1 4 n + 1 2 n 2 n . 1. ρ = 2 2. ρ = 1 16 3. ρ = 1 2 4. ρ = 1 4 5. ρ = 4 correct Explanation: After division, 4 n + 1 2 n = 4 + 1 /n 2 , so | a n | 1 /n = 4 + 1 /n 2 · 2 . On the other hand, lim n → ∞ 4 + 1 /n 2 = 2 . Consequently, ρ = 4 . keywords: 004 (part 1 of 1) 10 points Which of the following properties does the series X n = 1 ( - 5) n (2 n )! have? 1. divergent 2. absolutely convergent correct 3. conditionally convergent Explanation: The given series can be written as X n = 1 a n with a n defined by a n = ( - 5) n (2 n )! = ( - 5 n ) 1 · 2 · 3 . . . (2 n - 1) · 2 n . In this case, fl fl fl a n +1 a n fl fl fl = 5 (2 n + 1)(2 n + 2) -→ 0 as n → ∞ . Consequently, by the Ratio Test, the given series is absolutely convergent . keywords: 005 (part 1 of 1) 10 points Determine whether the series X k = 1 4 6 + 5 k converges or diverges. 1. series is convergent correct 2. series is divergent Explanation: Note first that the inequalities 0 < 4 6 + 5 k < 4 5 k
Su, Yung – Final 1 – Due: Dec 14 2007, 11:00 pm – Inst: Shinko Harper 3 hold for all n 1. On the other hand, the series X k = 1 4 5 k converges because it is a geometric series with | r | = 1 5 < 1 . By the comparison test, therefore, the series is convergent . keywords: 006 (part 1 of 1) 10 points Determine whether the series X n =0 2 7 n/ 2 is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 2 7 - 2 2. convergent with sum = 7 - 2 2 3. convergent with sum = 7 - 2 7 4. convergent with sum = 7 7 - 2 cor- rect 5. divergent Explanation: The infinite series X n =0 2 7 n/ 2 is an infinite geometric series n =0 ar n with a = 1 and r = 2 /

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Final 01 - Su Yung Final 1 Due 11:00 pm Inst Shinko Harper...

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