HW 1-solutions - husain (aih243) HW 1 mann (54675) 1 This...

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Unformatted text preview: husain (aih243) HW 1 mann (54675) 1 This print-out should have 23 questions. Multiple-choice 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.

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HW 1-solutions - husain (aih243) HW 1 mann (54675) 1 This...

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