FINAL - Martinez, Gorge Final 1 Due: Dec 14 2007, 10:00 pm...

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Martinez, Gorge – Final 1 – Due: Dec 14 2007, 10:00 pm – Inst: Diane Radin 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points Find the value o± lim x 0 e 5 x - e - 5 x sin 6 x . 1. limit = 11 6 2. limit does not exist 3. limit = 5 3 correct 4. limit = 10 7 5. limit = 2 6. limit = 5 4 Explanation: Set f ( x ) = e 5 x - e - 5 x , g ( x ) = sin 6 x. Then f, g are everywhere di²erentiable ±unc- 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 ) = 5( e 5 x + e - 5 x ) , g 0 ( x ) = 6 cos 6 x, while lim x 0 f 0 ( x ) = 10 , lim x 0 g 0 ( x ) = 6 . Consequently, lim x 0 e 5 x - e - 5 x sin 6 x = 5 3 . keywords: 002 (part 1 o± 1) 10 points Determine whether the sequence { a n } con- verges or diverges when a n = ( - 1) n µ 3 n + 4 3 n + 3 , and i± it does, fnd its limit. 1. limit = 0 2. limit = 1 3. sequence diverges correct 4. limit = 4 3 5. limit = ± 1 Explanation: A±ter division, 3 n + 4 3 n + 3 = 3 + 4 n 3 + 3 n . Now 4 n , 3 n 0 as n → ∞ , so lim n → ∞ 3 n + 4 3 n + 3 = 1 6 = 0 . Thus as n → ∞ , the values o± a n oscillate be- tween values ever closer to ± 1. Consequently, the sequence diverges . keywords: 003 (part 1 o± 1) 10 points
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Martinez, Gorge – Final 1 – Due: Dec 14 2007, 10:00 pm – Inst: Diane Radin 2 To apply the root test to an infnite series n a n the value o± ρ = lim n → ∞ | a n | 1 /n has to be determined. Compute the value o± ρ ±or the series X n = 1 µ 6 n + 1 5 n 2 n . 1. ρ = 36 25 correct 2. ρ = 6 5 3. ρ = 1 25 4. ρ = 1 5 5. ρ = 1 36 Explanation: A±ter division, 6 n + 1 5 n = 6 + 1 /n 5 , so | a n | 1 /n = 6 + 1 /n 5 · 2 . On the other hand, lim n → ∞ 6 + 1 /n 5 = 6 5 . Consequently, ρ = 36 25 . keywords: 004 (part 1 o± 1) 10 points Which o± the ±ollowing properties does the series X n = 1 ( - 4) n (2 n )! have? 1. conditionally convergent 2. divergent 3. absolutely convergent correct Explanation: The given series can be written as X n = 1 a n with a n defned by a n = ( - 4) n (2 n )! = ( - 4 n ) 1 · 2 · 3 . . . (2 n - 1) · 2 n . In this case, f f f a n +1 a n f f f = 4 (2 n + 1)(2 n + 2) -→ 0 as n → ∞ . Consequently, by the Ratio Test, the given series is absolutely convergent . keywords: 005 (part 1 o± 1) 10 points Determine whether the series X k = 1 4 6 + 2 k converges or diverges. 1. series is divergent 2. series is convergent correct Explanation: Note frst that the inequalities 0 < 4 6 + 2 k < 4 2 k
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Martinez, Gorge – Final 1 – Due: Dec 14 2007, 10:00 pm – Inst: Diane Radin 3 hold for all n 1. On the other hand, the series X k = 1 4 2 k converges because it is a geometric series with | r | = 1 2 < 1 . By the comparison test, therefore, the
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FINAL - Martinez, Gorge Final 1 Due: Dec 14 2007, 10:00 pm...

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