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HW 1-solutions

# HW 1-solutions - morris(gmm643 HW 1 mann(54675 This...

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morris (gmm643) – 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. none of them correct 2. A and B only 3. B only 4. A only 5. B and C only 6. C only 7. all of them 8. A and C only Explanation: A. By properties of limits lim x 1 g ( x ) G ( x ) = 0 = 0 , so this limit is not an indeterminate form. B. By properties of limits lim x 1 f ( x ) g ( x ) = 0 · 0 = 0 , so this limit is not an indeterminate form. C. By properties of limits lim x 1 F ( x ) g ( x ) = 2 0 = 1 , so this limit is not an indeterminate form. 002 10.0 points Determine the value of lim x → ∞ x x 2 + 7 . 1. limit = 1 2 2. limit = 3. limit = 1 4 4. limit = 4 5. limit = 2 6. limit = 1 correct 7. limit = 0 Explanation: Since lim x →∞ x x 2 + 7 , 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 ( x ) g ( x ) with f ( x ) = x , g ( x ) = radicalbig x 2 + 7 to evaluate the limit. But f ( x ) = 1 , g ( x ) = x x 2 + 7 ,

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morris (gmm643) – HW 1 – mann – (54675) 2 so lim x → ∞ f ( x ) g ( x ) = lim x → ∞ x 2 + 7 x = , which is again of indeterminate form. Let’s try using L’Hospital’s Rule again but now with f ( x ) = radicalbig x 2 + 7 , g ( x ) = x , and f ( x ) = x x 2 + 7 , g ( x ) = 1 . In this case, lim x → ∞ x 2 + 7 x = lim x → ∞ x x 2 + 7 , 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 + 7 = x | x | radicalbig 1 + 7 /x 2 = 1 radicalbig 1 + 7 /x 2 for x > 0. Thus lim x →∞ x x 2 + 7 = lim x → ∞ 1 radicalbig 1 + 7 /x 2 , and so limit = 1 . 003 10.0 points Determine if lim x 0 e 4 x 1 sin 5 x exists, and if it does, find its value. 1. none of the other answers 2. limit = 0 3. limit = 4 5 correct 4. limit = 5. limit = −∞ 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 0 f ( x ) = 0 , lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule can be applied: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) . But f ( x ) = 4 e 4 x , g ( x ) = 5 cos 5 x , and so lim x 0 f ( x ) = 4 , lim x 0 g ( x ) = 5 . Consequently, the limit exists and limit = 4 5 . 004 10.0 points Determine if lim x 5 parenleftBig 4 ln( x 4) 4 x 5 parenrightBig exists, and if it does, find its value. 1. limit = 0 2. limit = −∞ 3. limit = + 4. none of the other answers 5. limit = 4
morris (gmm643) – HW 1 – mann – (54675) 3 6. limit = 2 correct Explanation: After 4 ln( x 4) 4 x 5 is brought to a common denominator it can be written as f ( x ) g ( x ) = 4( x 5) 4 ln( x 4) ( x 5) ln( x 4) where f and g are functions which are differ- entiable on the interval (4 , ). In addition, lim x 5 f ( x ) = 0 , lim x 5 g ( x ) = 0 , so L’Hospital’s Rule applies. Thus lim x

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