hw 1 - rhodes (ajr2283) HW01 Gilbert (56380) 1 This...

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Unformatted text preview: rhodes (ajr2283) HW01 Gilbert (56380) 1 This print-out should have 20 questions. Multiple-choice 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 does not exist 2. limit = 7 2 correct 3. limit = 3 4. limit = 4 5. limit = 7 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 4 2. limit = 1 2 3. limit = 4. limit = 1 correct 5. limit = 2 6. limit = 0 7. limit = 4 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 , rhodes (ajr2283) 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 = 4 3 correct 2. limit =- 3. none of the other answers 4. limit = 5. limit = 3 4 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 g ( x ) G ( x ) ; C. lim x 1 f ( x ) g ( x ) ; are indeterminate forms?...
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hw 1 - rhodes (ajr2283) HW01 Gilbert (56380) 1 This...

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