{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Gilbert_FinalExamsol

# Gilbert_FinalExamsol - Version 067 FINAL Gilbert(56380 This...

This preview shows pages 1–4. Sign up to view the full content.

Version 067 – FINAL – Gilbert – (56380) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x 0 e 5 x 1 sin 3 x exists, and if it does, find its value. 1. limit = 2. none of the other answers 3. limit = 5 3 correct 4. limit = 0 5. limit = −∞ 6. limit = 3 5 Explanation: Set f ( x ) = e 5 x 1 , g ( x ) = sin 3 x . Then f and g are differentiable functions such that lim x 0 f ( x ) = 0 , lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule can be applied: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) . But f ( x ) = 5 e 5 x , g ( x ) = 3 cos 3 x , and so lim x 0 f ( x ) = 5 , lim x 0 g ( x ) = 3 . Consequently, the limit exists and limit = 5 3 . 002 10.0 points Determine if the improper integral I = integraldisplay 1 6 tan 1 x 1 + x 2 dx converges, and if it does, find its value. 1. I = 9 8 π 2 2. I does not converge 3. I = 3 4 4. I = 9 16 π 2 correct 5. I = 9 16 6. I = 3 4 π 2 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 6 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 = π 4 , and x = t = u = tan 1 t . In this case I t = integraldisplay tan - 1 t π/ 4 6 u du = 3 bracketleftBig u 2 bracketrightBig tan - 1 t π/ 4 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Version 067 – FINAL – Gilbert – (56380) 2 Thus I t = 3 braceleftBig ( tan 1 t ) 2 π 2 16 bracerightBig . On the other hand, lim t → ∞ tan 1 t = π 2 . Consequently, I converges and has value I = 3 parenleftBig 1 4 1 16 parenrightBig π 2 = 9 16 π 2 . 003 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = ln(5 n 6 ) ln(4 n 3 ) , and if it converges, find the limit. 1. converges with limit = 2 correct 2. converges with limit = 5 4 3. diverges 4. converges with limit = ln 5 ln 4 5. converges with limit = 0 Explanation: By properties of logs, ln(5 n 6 ) = ln 5 + 6 ln n , ln(4 n 3 ) = ln 4 + 3 ln n . Thus a n = ln 5 + 6 ln n ln 4 + 3 ln n = 6 + ln 5 ln n 3 + ln 4 ln n . On the other hand, lim n →∞ ln 5 ln n = lim n →∞ ln 4 ln n = 0 . Properties of limits thus ensure that the given sequence converges with limit = 2 . 004 10.0 points Which, if any, of the following series con- verge? A . summationdisplay n =1 n 1 + n 2 B . 1 + 1 4 + 1 9 + 1 16 + . . . 1. both of them 2. neither of them 3. B only correct 4. A only Explanation: A. Use f ( x ) = x 1 + x 2 . Then integraldisplay 1 f ( x ) dx is divergent (ln integral), so series di- verges. B. Series is summationdisplay n =1 1 n 2 . Use f ( x ) = 1 x 2 . Then integraldisplay 1 f ( x ) dx is convergent, so series converges. 005 10.0 points Which of the following series converge?
Version 067 – FINAL – Gilbert – (56380) 3 A. summationdisplay k =1 parenleftBig 3 k 5 k + 1 parenrightBig k B.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 14

Gilbert_FinalExamsol - Version 067 FINAL Gilbert(56380 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online