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Gilbert_ReviewExam01sol

# Gilbert_ReviewExam01sol - moseley(cmm3869 ReviewExam01...

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moseley (cmm3869) – ReviewExam01 – Gilbert – (56380) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 0.0 points When f, g, F and G are functions such that lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = , lim x 1 F ( x ) = 2 , lim x 1 G ( x ) = , which, if any, of A. lim x 1 g ( x ) f ( x ) , B. lim x 1 f ( x ) F ( x ) , C. lim x 1 G ( x ) g ( x ) , are indeterminate forms? 1. B and C only 2. A and C only 3. none of them 4. A and B only 5. B only 6. C only correct 7. all of them 8. A only Explanation: A. By properties of limits, lim x 1 g ( x ) f ( x ) = 0 = , so this limit is not an indeterminate form. B. By properties of limits, lim x 1 f ( x ) F ( x ) = 0 2 = 0 , so this limit is not an indeterminate form. C. Since lim x 1 G ( x ) g ( x ) = ∞ − ∞ , this limit is an indeterminate form. 002 0.0 points Determine lim x 1 e 5 x e 5 ln(3 x 2) . 1. limit = 5 3 e 5 correct 2. limit = e 5 3. limit = 5 4 e 5 4. limit does not exist 5. limit = 5 3 6. limit = e 5 3 Explanation: Since e 5 x e 5 ln(3 x 14) = f ( x ) g ( x ) where f and g are differentiable on (0 , ) and lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = 0 , L’Hospital’s Rule can be applied. Now f ( x ) = 5 e 5 x , g ( x ) = 3 3 x 2 , so lim x 1 f ( x ) = 5 e 5 , lim x 1 g ( x ) = 3 .

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moseley (cmm3869) – ReviewExam01 – Gilbert – (56380) 2 Consequently, lim x 1 e 5 x e 5 ln(3 x 2) = 5 3 e 5 . 003 0.0 points Find the value of lim x 0 1 cos x 5 sin 2 3 x . 1. limit does not exist 2. limit = 1 90 correct 3. limit = 1 45 4. limit = 1 30 5. limit = 2 45 Explanation: Set f ( x ) = 1 cos x, g ( x ) = 5 sin 2 3 x . Then f, g are differentiable functions such that lim x 0 f ( x ) = 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 ) = lim x 0 sin x 30 sin 3 x cos 3 x . To compute this last limit we can either apply L’Hospital’s Rule again or use the fact that lim x 0 sin x x = 1 , lim x 0 sin 3 x x = 3 . Consequently, lim x 0 1 cos x 5 sin 2 3 x = 1 90 . 004 0.0 points Determine if the improper integral I = integraldisplay 1 0 4 ln 3 x dx converges, and if it does, find its value. 1. I = 3(ln 4 1) 2. I = 4 ln 3 3 3. I = 4(ln 3 + 1) 4. I does not converge 5. I = 4(ln 3 1) correct 6. I = 3 ln 4 + 4 7. I = 3(ln 4 + 1) Explanation: Since ln 3 x → −∞ as x 0 through pos- itive values of x , the graph of 4 ln 3 x has a vertical asymptote at x = 0. It is this that makes I an improper integral. So we set I = lim t 0+ integraldisplay 1 t 4 ln 3 x dx and check if the limit exists.
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Gilbert_ReviewExam01sol - moseley(cmm3869 ReviewExam01...

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