This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 037 FINAL Gilbert (56380) 1 This printout should have 25 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine if lim x e 3 x 1 sin2 x exists, and if it does, find its value. 1. limit = 0 2. limit = 2 3 3. limit = 4. none of the other answers 5. limit = 6. limit = 3 2 correct Explanation: Set f ( x ) = e 3 x 1 , g ( x ) = sin 2 x. Then f and g are differentiable functions such that lim x f ( x ) = 0 , lim x g ( x ) = 0 . Thus LHospitals Rule can be applied: lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) . But f ( x ) = 3 e 3 x , g ( x ) = 2 cos 2 x, and so lim x f ( x ) = 3 , lim x g ( x ) = 2 . Consequently, the limit exists and limit = 3 2 . 002 10.0 points Determine if the improper integral I = integraldisplay 1 tan 1 x 1 + x 2 dx converges, and if it does, find its value. 1. I = 3 16 2 2. I = 3 32 3. I = 1 8 2 4. I = 3 32 2 correct 5. I = 1 8 6. I does not converge Explanation: The integral I is improper because the in terval of integration is infinite. Thus the integral will converge if lim t I t , I t = integraldisplay t 1 tan 1 x 1 + x 2 dx, exists, and its value will then be the value of the limit. The substitution u = tan 1 x is suggested. For then du = 1 1 + x 2 dx, while x = 1 = u = 4 , and x = t = u = tan 1 t . In this case I t = integraldisplay tan 1 t / 4 udu = 1 2 bracketleftBig u 2 bracketrightBig tan 1 t / 4 . Version 037 FINAL Gilbert (56380) 2 Thus I t = 1 2 braceleftBig ( tan 1 t ) 2 2 16 bracerightBig . On the other hand, lim t tan 1 t = 2 . Consequently, I converges and has value I = 1 2 parenleftBig 1 4 1 16 parenrightBig 2 = 3 32 2 . 003 10.0 points Determine whether the sequence { a n } con verges or diverges when a n = ln(4 n 3 ) ln(3 n 4 ) , and if it converges, find the limit. 1. converges with limit = 4 3 2. converges with limit = 0 3. diverges 4. converges with limit = ln 4 ln 3 5. converges with limit = 3 4 correct Explanation: By properties of logs, ln(4 n 3 ) = ln 4 + 3 ln n , ln(3 n 4 ) = ln 3 + 4 ln n . Thus a n = ln4 + 3 ln n ln3 + 4 ln n = 3 + ln 4 ln n 4 + ln 3 ln n . On the other hand, lim n ln 4 ln n = lim n ln 3 ln n = 0 . Properties of limits thus ensure that the given sequence converges with limit = 3 4 . 004 10.0 points Which, if any, of the following series con verge? A . summationdisplay n =1 1 1 + n 2 B . 1 + 1 4 + 1 9 + 1 16 + . . . 1. B only 2. both of them correct 3. A only 4. neither of them Explanation: A. Use f ( x ) = 1 1 + x 2 . Then integraldisplay 1 f ( x ) dx is convergent (tan 1 integral), so series converges....
View Full
Document
 Spring '07
 TextbookAnswers
 Multivariable Calculus

Click to edit the document details