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HW 1-solutions

HW 1-solutions - husain(aih243 HW 1 mann(54675 This...

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husain (aih243) – HW 1 – mann – (54675) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd 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. A and C only correct 2. B and C only 3. C only 4. none oF them 5. all oF them 6. A only 7. A and B only 8. B only Explanation: A. Since lim x 1 = ∞ · 0 , this limit is an indeterminate Form. B. By properties oF limits lim x 1 F ( x ) g ( x ) = 2 0 = 1 , so this limit is not an indeterminate Form. C. Since lim x 1 f ( x ) g ( x ) = 0 0 , this limit is an indeterminate Form. 002 10.0 points Determine the value oF lim x →∞ x x 2 + 5 . 1. limit = 2. limit = 1 4 3. limit = 4 4. limit = 0 5. limit = 1 correct 6. limit = 1 2 7. limit = 2 Explanation: Since lim x →∞ x x 2 + 5 , the limit is oF indeterminate Form. We might frst 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 ) = r x 2 + 5 to evaluate the limit. But f ( x ) = 1 , g ( x ) = x x 2 + 5 ,
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husain (aih243) – HW 1 – mann – (54675) 2 so lim x →∞ f ( x ) g ( x ) = lim x x 2 + 5 x = , which is again of indeterminate form. Let’s try using L’Hospital’s Rule again but now with f ( x ) = r x 2 + 5 , g ( x ) = x , and f ( x ) = x x 2 + 5 , g ( x ) = 1 . In this case, lim x x 2 + 5 x = lim x x x 2 + 5 , 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 + 5 = x | x | r 1 + 5 /x 2 = 1 r 1 + 5 /x 2 for x > 0. Thus lim x →∞ x x 2 + 5 = lim x 1 r 1 + 5 /x 2 , and so limit = 1 . 003 10.0 points Determine if lim x 0 e 4 x 1 sin5 x exists, and if it does, Fnd its value. 1. limit = 2. none of the other answers 3. limit = 0 4. limit = −∞ 5. limit = 4 5 correct 6. limit = 5 4 Explanation: Set f ( x ) = e 4 x 1 , g ( x ) = sin 5 x . Then f and g are di±erentiable 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 cos5 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 p 1 ln( x 4) 1 x 5 P exists, and if it does, Fnd its value. 1. limit = 1 2. limit = 0 3. limit = −∞ 4. none of the other answers 5. limit = +
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husain (aih243) – HW 1 – mann – (54675) 3 6. limit = 1 2 correct Explanation: After 1 ln( x 4) 1 x 5 is brought to a common denominator it can be written as f ( x ) g ( x ) = ( x 5) ln( x 4) ( x 5) ln( x 4) where f and g are functions which are diFer- 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 5 f ( x ) g ( x ) = lim x 5 f ( x ) g ( x ) with f ( x ) = 1 1 x 4 = ( x 5) x 4 , g ( x ) = ln( x 4) + x 5 x 4 .
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HW 1-solutions - husain(aih243 HW 1 mann(54675 This...

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