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Unformatted text preview: mao (tm23477) Hw 15 gualdani (57180) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Determine if lim x 1 parenleftBig 3 x 2 + 1 x 2 + 1 parenrightBig exists, and if it does, find its value. 1. limit does not exist 2. limit = 3 3. limit = 4 4. limit = 1 5. limit = 2 correct Explanation: Set f ( x ) = 3 x 2 + 1 , g ( x ) = x 2 + 1 . Then lim x 1 f ( x ) = 4 , lim x 1 g ( x ) = 2 . Thus the limits for both the numerator and denominator exist and neither is zero; so LHospitals rule does not apply. In fact, all we have to do is use properties of limits. For then we see that limit = 2 . 002 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 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. B and C only 2. none of them correct 3. A and C only 4. B only 5. A and B only 6. A only 7. C only 8. all of them Explanation: A. Since lim x 1 f ( x ) g ( x ) = , this limit is an indeterminate form. B. Since lim x 1 f ( x ) g ( x ) = 0 , this limit is an indeterminate form. C. Since lim x 1 = , this limit is an indeterminate form. 003 10.0 points Determine the value of lim x x x 2 + 1 . 1. limit = 1 correct mao (tm23477) Hw 15 gualdani (57180) 2 2. limit = 4 3. limit = 4. limit = 2 5. limit = 1 4 6. limit = 1 2 7. limit = 0 Explanation: Since lim x x x 2 + 1 , 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 + 1 to evaluate the limit. But f ( x ) = 1 , g ( x ) = x x 2 + 1 , so lim x f ( x ) g ( x ) = lim x x 2 + 1 x = , which is again of indeterminate form. Lets try using LHospitals Rule again but now with f ( x ) = radicalbig x 2 + 1 , g ( x ) = x , and f ( x ) = x x 2 + 1 , g ( x ) = 1 . In this case, lim x x 2 + 1 x = lim x x x 2 + 1 , 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 + 1 = x  x  radicalbig 1 + 1 /x 2 = 1 radicalbig 1 + 1 /x 2 for x > 0. Thus lim x x x 2 + 1 = lim x 1 radicalbig 1 + 1 /x 2 , and so limit = 1 . 004 10.0 points Determine if the limit lim x 1 x 2 1 x 3 1 exists, and if it does, find its value....
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 Fall '08
 schultz

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