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Unformatted text preview: husain (aih243) HW 1 mann (54675) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Sections 7.8, 8.8, and 12.1 001 10.0 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 g ( x ) G ( x ) , B. lim x 1 F ( x ) g ( x ) , C. lim x 1 f ( x ) g ( x ) , are indeterminate forms? 1. A and C only correct 2. B and C only 3. C only 4. none of them 5. all of them 6. A only 7. A and B only 8. B only Explanation: A. Since lim x 1 = , 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 f ( x ) g ( x ) = , this limit is an indeterminate form. 002 10.0 points Determine the value of lim x x x 2 + 5 . 1. limit = 2. limit = 1 4 3. limit = 4 4. limit = 0 5. limit = 1 correct 6. limit = 1 2 7. limit = 2 Explanation: Since lim x x x 2 + 5 , the limit is of indeterminate form. We might first try to use LHospitals Rule lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) with f ( x ) = x , g ( x ) = radicalbig x 2 + 5 to evaluate the limit. But f ( x ) = 1 , g ( x ) = x x 2 + 5 , husain (aih243) HW 1 mann (54675) 2 so lim x f ( x ) g ( x ) = lim x x 2 + 5 x = , which is again of indeterminate form. Lets try using LHospitals Rule again but now with f ( x ) = radicalbig x 2 + 5 , g ( x ) = x , and f ( x ) = x x 2 + 5 , g ( x ) = 1 . In this case, lim x x 2 + 5 x = lim x x x 2 + 5 , which is the limit we started with. So, this is an example where LHospitals Rule applies, but doesnt work! We have to go back to algebraic methods: x x 2 + 5 = x  x  radicalbig 1 + 5 /x 2 = 1 radicalbig 1 + 5 /x 2 for x > 0. Thus lim x x x 2 + 5 = lim x 1 radicalbig 1 + 5 /x 2 , and so limit = 1 . 003 10.0 points Determine if lim x e 4 x 1 sin5 x exists, and if it does, find its value. 1. limit = 2. none of the other answers 3. limit = 0 4. limit = 5. limit = 4 5 correct 6. limit = 5 4 Explanation: Set f ( x ) = e 4 x 1 , g ( x ) = sin 5 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 ) = 4 e 4 x , g ( x ) = 5 cos5 x , and so lim x f ( x ) = 4 , lim x g ( x ) = 5 ....
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This note was uploaded on 11/30/2010 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Sadler
 Limits

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