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Unformatted text preview: byrne (cmb3744) – HW 01 – rusin – (55565) 1 This printout should have 15 questions. Multiplechoice 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 10.0 points Determine if the limit lim x → 1 x 8 + 1 x 5 + 1 exists, and if it does, find its value. 1. limit = 0 2. limit = 1 correct 3. limit = 8 5 4. limit does not exist 5. limit = 5 8 Explanation: Set f ( x ) = x 8 + 1 , g ( x ) = x 5 + 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 10.0 points Determine if the limit lim x → 1 x 8 1 x 3 1 exists, and if it does, find its value. 1. limit = ∞ 2. limit =∞ 3. none of the other answers 4. limit = 3 8 5. limit = 8 3 correct Explanation: Set f ( x ) = x 8 1 , g ( x ) = x 3 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 ) = 8 x 7 , g ′ ( x ) = 3 x 2 . Consequently, limit = 8 3 . 003 10.0 points Determine the value of lim x →∞ x √ x 2 + 1 . 1. limit = 1 4 2. limit = 2 3. limit = 4 4. limit = 1 2 byrne (cmb3744) – HW 01 – rusin – (55565) 2 5. limit = ∞ 6. limit = 0 7. limit = 1 correct 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 ′ ( x ) g ′ ( x ) with f ( x ) = x , g ( x ) = radicalbig x 2 + 1 to evaluate the limit. But f ′ ( x ) = 1 , g ′ ( x ) = x √ x 2 + 1 , so lim x →∞ f ′ ( x ) g ′ ( 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 ) = radicalbig x 2 + 1 , g ( x ) = x , and f ′ ( x ) = x √ x 2 + 1 , g ′ ( 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  radicalbig 1 + 1 /x 2 = 1 radicalbig 1 + 1 /x 2 for x > 0. Thus lim x →∞ x √ x 2 + 1 = lim x →∞ 1 radicalbig 1 + 1 /x 2 , and so limit = 1 . 004 10.0 points Find the value of lim x → e 4 x e − 4 x sin 6 x ....
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This note was uploaded on 05/08/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas.
 Spring '09
 Chu
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