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# HW1 - hussain(tsh476 HW 01 rusin(55565 This print-out...

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hussain (tsh476) – HW 01 – rusin – (55565) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Welcome to the Quest homework system. This is what most homework sets will be like: you have one week to finish them, which will give you two chances to get help when you meet with the TA, if you start the problems by the weekend. Please inform me or the TA if there are any glitches in the system. 001 10.0points Determine if the limit lim x 1 x 8 + 1 x 9 + 1 exists, and if it does, find its value. 1. limit = 8 9 2. limit = 0 3. limit does not exist 4. limit = 1 correct 5. limit = 9 8 Explanation: Set f ( x ) = x 8 + 1 , g ( x ) = x 9 + 1 . Then lim x 1 f ( x ) = lim x 1 g ( x ) = 1 , so L’Hospital’s rule does not apply. In fact, by properties of limits we see that limit = 1 . 002 10.0points Determine if the limit lim x 1 x 2 - 1 x 5 - 1 exists, and if it does, find its value. 1. limit = 2. limit = 2 5 correct 3. limit = -∞ 4. none of the other answers 5. limit = 5 2 Explanation: Set f ( x ) = x 2 - 1 , g ( x ) = x 5 - 1 . Then lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = 0 , so L’Hospital’s rule applies. Thus lim x 1 f ( x ) g ( x ) = lim x 1 f ( x ) g ( x ) . But f ( x ) = 2 x 1 , g ( x ) = 5 x 4 . Consequently, limit = 2 5 . 003 10.0points Determine the value of lim x → ∞ x x 2 + 7 . 1. limit = 1 2 2. limit = 1 correct 3. limit = 4. limit = 1 4

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hussain (tsh476) – HW 01 – rusin – (55565) 2 5. limit = 4 6. limit = 0 7. limit = 2 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 , 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 . 004 10.0points Find the value of lim x 0 e 3 x - e 3 x sin 5 x . 1. limit = 8 5 2. limit = 1 3. limit = 6 7 4. limit = 7 5 5. limit does not exist 6. limit = 6 5 correct Explanation: Set f ( x ) = e 3 x - e 3 x , g ( x ) = sin 5 x. Then f, g are everywhere differentiable func- tions such that lim x 0 f ( x ) = lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule applies: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) . Now f ( x ) = 3( e 3 x + e 3 x ) , g ( x ) = 5 cos 5 x, while lim x 0 f ( x ) = 6 , lim x 0 g ( x ) = 5 .
hussain (tsh476) – HW 01 – rusin – (55565) 3 Consequently, lim x 0 e 3 x - e 3 x sin 5 x = 6 5 .

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HW1 - hussain(tsh476 HW 01 rusin(55565 This print-out...

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