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Unformatted text preview: Barkley, Thane Review 3 Due: Dec 11 2007, 6:00 pm Inst: Louiza Fouli 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. Recall that a sphere of radius r has volume = 4 3 r 3 , surface area = 4 r 2 . 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 g ( x ) G ( x ) ; C. lim x 1 f ( x ) g ( x ) ; are NOT indeterminate forms? 1. B only 2. A and C only 3. B and C only 4. C only 5. A and B only 6. A only 7. all of them correct 8. none of them Explanation: A. By properties of limits lim x 1 F ( x ) g ( x ) = 2 = 1 , so this limit is not an indeterminate form. B. By properties of limits lim x 1 g ( x ) G ( x ) = = 0 , so this limit is not an indeterminate form. C. By properties of limits lim x 1 f ( x ) g ( x ) = 0 0 = 0 , so this limit is not an indeterminate form. keywords: 002 (part 1 of 1) 10 points Determine if the limit lim x 1 x 2 1 x 7 1 exists, and if it does, find its value. 1. none of the other answers 2. limit = 3. limit = 7 2 4. limit = 2 7 correct 5. limit = Explanation: Set f ( x ) = x 2 1 , g ( x ) = x 7 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 ) . Barkley, Thane Review 3 Due: Dec 11 2007, 6:00 pm Inst: Louiza Fouli 2 But f ( x ) = 2 x 1 , g ( x ) = 7 x 6 . Consequently, limit = 2 7 . 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 = 1 5 2. limit = 0 correct 3. limit = 4. limit = 5. none of the other answers 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 2 e 2 x + 3 e 2 x e 2 x 2 e 2 x . 1. limit = 1 2 2. limit = 1 3 3. limit = 1 3 4. limit = 1 2 5. limit = 2 6. limit = 2 correct Explanation: After division we see that 2 e 2 x + 3 e 2 x e 2 x 2 e 2 x = 2 + 3 e 4 x 1 2 e 4 x . On the other hand, lim x e ax = 0 for all a > 0. But then by properties of limits, lim x 2 + 3 e 4 x 1 2 e 4 x = 2 ....
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This note was uploaded on 04/16/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz
 Calculus

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