review 3 ans - Lescure, Etienne Review 3 Due: Dec 10 2007,...

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Unformatted text preview: Lescure, Etienne Review 3 Due: Dec 10 2007, 11:00 am Inst: Diane Radin 1 This print-out should have 18 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 When f, g, F and G are functions such that lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = 0 , lim x 1 F ( x ) = 2 , lim x 1 G ( x ) = , which, if any, of A. lim x 1 f ( x ) g ( x ) ; B. lim x 1 F ( x ) g ( x ) ; C. lim x 1 g ( x ) G ( x ) ; are NOT indeterminate forms? 1. A only 2. B only correct 3. none of them 4. B and C only 5. A and C only 6. all of them 7. C only 8. A and B only Explanation: A. Since lim x 1 f ( x ) g ( x ) = , this limit is an indeterminate form. B. By properties of limits lim x 1 F ( x ) g ( x ) = 2 = 1 , so this limit is not an indeterminate form. C. Since lim x 1 = , this limit is an indeterminate form. keywords: 002 (part 1 of 1) 10 points Determine if the limit lim x 1 x 8- 1 x 3- 1 exists, and if it does, find its value. 1. none of the other answers 2. limit = 8 3 correct 3. limit = 4. limit =- 5. limit = 3 8 Explanation: Set f ( x ) = x 8- 1 , g ( x ) = x 3- 1 . Then lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = 0 , so LHospitals rule applies. Thus lim x 1 f ( x ) g ( x ) = lim x 1 f ( x ) g ( x ) . But f ( x ) = 8 x 7 , g ( x ) = 3 x 2 . Consequently, limit = 8 3 . Lescure, Etienne Review 3 Due: Dec 10 2007, 11:00 am Inst: Diane Radin 2 keywords: LHospitals rule, rational func- tion, zero over zero 003 (part 1 of 1) 10 points Evaluate lim x x 2 e 5 x . 1. limit = 2. limit =- 3. limit = 0 correct 4. none of the other answers 5. limit = 1 5 6. limit = 1 Explanation: Since lim x x 2 e 5 x , the limit is of indeterminate form, so we apply LHospitals Rule: lim x x 2 e 5 x = lim x 2 x 5 e 5 x = . Applying LHospitals Rule once again, there- fore, we arrive at lim x x 2 e 5 x = lim x 2 25 e 5 x = 0 . keywords: 004 (part 1 of 1) 10 points Find the value of lim x 4 e 3 x + e- 3 x 2 e 3 x- 3 e- 3 x . 1. limit =- 2 2. limit = 2 correct 3. limit = 1 2 4. limit =- 1 2 5. limit =- 3 5 6. limit = 3 5 Explanation: After division we see that 4 e 3 x + e- 3 x 2 e 3 x- 3 e- 3 x = 4 + e- 6 x 2- 3 e- 6 x . On the other hand, lim x e- ax = 0 for all a > 0. But then by properties of limits, lim x 4 + e- 6 x 2- 3 e- 6 x = 2 . Consequently, limit = 2 ....
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review 3 ans - Lescure, Etienne Review 3 Due: Dec 10 2007,...

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