HW 1-solutions - morris (gmm643) – HW 1 – mann –...

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Unformatted text preview: 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 , 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 = 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 , 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 → e 4 x − 1 sin5 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 → f ( x ) = 0 , lim x → g ( x ) = 0 . Thus L’Hospital’s Rule can be applied: lim x → f ( x ) g ( x ) = lim x → f ′ ( x ) g ′ ( x ) ....
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This note was uploaded on 11/17/2010 for the course PHY 56630 taught by Professor Coker during the Spring '10 term at University of Texas at Austin.

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HW 1-solutions - morris (gmm643) – HW 1 – mann –...

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