# hw15 - Granillo Yvette Homework 15 Due Dec 9 2005 3:00 am...

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Granillo, Yvette – Homework 15 – Due: Dec 9 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points When f, g, F and G are functions such that lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = , 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 g ( x ) f ( x ) , C. lim x 1 f ( x ) F ( x ) g ( x ) , are indeterminate forms? 1. B and C only 2. all of them 3. B only 4. A and B only 5. C only 6. A only correct 7. A and C only 8. none of them Explanation: A. Since lim x 1 g ( x ) G ( x ) = , this limit is an indeterminate form. B. By properties of limits, lim x 1 g ( x ) f ( x ) = 0 = , so this limit is not an indeterminate form. C. By properties of limits, lim x 1 f ( x ) F ( x ) g ( x ) = 0 · 2 = 0 , so this limit is not an indeterminate form. keywords: Stewart5e, 002 (part 1 of 1) 10 points When f, g, F and G are functions such that lim x 2 f ( x ) = 0 , lim x 2 g ( x ) = 1 , lim x 2 F ( x ) = 2 , lim x 2 G ( x ) = , which of the following is an indeterminate form? 1. lim x 2 F ( x ) G ( x ) 2. lim x 2 F ( x ) f ( x ) 3. lim x 2 g ( x ) G ( x ) correct 4. lim x 2 f ( x ) g ( x ) 5. lim x 2 g ( x ) F ( x ) Explanation: Since lim x 2 g ( x ) G ( x ) = 1 , lim x 2 F ( x ) G ( x ) = 2 , lim x 2 f ( x ) g ( x ) = 0 1 , lim x 2 g ( x ) F ( x ) = 1 2 , lim x 2 F ( x ) f ( x ) = 2 0 ,

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Granillo, Yvette – Homework 15 – Due: Dec 9 2005, 3:00 am – Inst: Edward Odell 2 we see that only lim x 2 g ( x ) G ( x ) is an indeterminate form. keywords: Stewart5e, 003 (part 1 of 1) 10 points Determine the value of lim x → ∞ x x 2 + 1 . 1. limit = 4 2. limit = 1 4 3. limit = 1 correct 4. limit = 1 2 5. limit = 0 6. limit = 2 7. limit = Explanation: Since lim x →∞ x x 2 + 1 , the limit is of indeterminate form. We might first try to use L’Hospital’s Rule lim x → ∞ f ( x ) g ( x ) = lim x → ∞ f 0 ( x ) g 0 ( x ) with f ( x ) = x , g ( x ) = p x 2 + 1 to evaluate the limit. But f 0 ( x ) = 1 , g 0 ( x ) = x x 2 + 1 , so lim x → ∞ f 0 ( x ) g 0 ( x ) = lim x → ∞ x 2 + 1 x = , which is again of indeterminate form. Let’s try using L’Hospital’s Rule again but now with f ( x ) = p x 2 + 1 , g ( x ) = x , and f 0 ( x ) = x x 2 + 1 , g 0 ( 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 L’Hospital’s Rule applies, but doesn’t work! We have to go back to algebraic methods: x x 2 + 1 = x | x | p 1 + 1 /x 2 = 1 p 1 + 1 /x 2 for x > 0. Thus lim x →∞ x x 2 + 1 = lim x → ∞ 1 p 1 + 1 /x 2 , and so limit = 1 . 004 (part 1 of 1) 10 points Determine if the limit lim x 1 x 2 - 1 x 9 - 1 exists, and if it does, find its value. 1. limit does not exist 2. limit = 9 2 3. limit = -∞ 4. limit = 2 9 correct
Granillo, Yvette – Homework 15 – Due: Dec 9 2005, 3:00 am – Inst: Edward Odell 3 5. limit = Explanation: Set f ( x ) = x 2 - 1 , g ( x ) = x 9 - 1 .

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