This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Su, Yung Final 1 Due: Dec 14 2007, 11:00 pm Inst: Shinko Harper 1 This printout should have 25 questions. Multiplechoice 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 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 ) = sin4 x. Then f, g are everywhere differentiable func tions such that lim x f ( x ) = lim x g ( x ) = 0 . Thus LHospitals Rule applies: lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) . Now f ( x ) = 2( e 2 x + e 2 x ) , g ( x ) = 4 cos 4 x, while lim x f ( x ) = 4 , lim x g ( x ) = 4 . Consequently, lim x 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 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 = lim n  a n  1 /n has to be determined. Compute the value of for the series X n = 1 4 n + 1 2 n 2 n . 1. = 2 2. = 1 16 3. = 1 2 4. = 1 4 5. = 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, = 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) 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 < 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 ....
View
Full
Document
This note was uploaded on 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Harper
 Calculus

Click to edit the document details