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Unformatted text preview: moseley (cmm3869) HW01 Gilbert (56380) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Welcome to Quest. Print off this as signment and bring it to lectures as well as discussion groups. 001 10.0 points Determine if lim x 1 parenleftBig 4 x 2 + 3 x 2 + 1 parenrightBig exists, and if it does, find its value. 1. limit = 7 2 correct 2. limit = 3 3. limit does not exist 4. limit = 7 5. limit = 4 Explanation: Set f ( x ) = 4 x 2 + 3 , g ( x ) = x 2 + 1 . Then lim x 1 f ( x ) = 7 , 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 = 7 2 . 002 10.0 points Determine the value of lim x x x 2 + 6 . 1. limit = 1 correct 2. limit = 0 3. limit = 1 4 4. limit = 4 5. limit = 6. limit = 2 7. limit = 1 2 Explanation: Since lim x x x 2 + 6 , 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 + 6 to evaluate the limit. But f ( x ) = 1 , g ( x ) = x x 2 + 6 , so lim x f ( x ) g ( x ) = lim x x 2 + 6 x = , which is again of indeterminate form. Lets try using LHospitals Rule again but now with f ( x ) = radicalbig x 2 + 6 , g ( x ) = x , and f ( x ) = x x 2 + 6 , g ( x ) = 1 . In this case, lim x x 2 + 6 x = lim x x x 2 + 6 , moseley (cmm3869) HW01 Gilbert (56380) 2 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 + 6 = x  x  radicalbig 1 + 6 /x 2 = 1 radicalbig 1 + 6 /x 2 for x > 0. Thus lim x x x 2 + 6 = lim x 1 radicalbig 1 + 6 /x 2 , and so limit = 1 . 003 10.0 points Determine if the limit lim x 1 x 4 1 x 3 1 exists, and if it does, find its value. 1. limit = 3 4 2. limit = 4 3 correct 3. limit = 4. none of the other answers 5. limit = Explanation: Set f ( x ) = x 4 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 ) = 4 x 3 , g ( x ) = 3 x 2 . Consequently, limit = 4 3 . 004 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 indeterminate forms?...
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 Spring '07
 Sadler
 Multivariable Calculus

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