Review Exam 1 - rhodes(ajr2283 – ReviewExam01 – Gilbert...

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Unformatted text preview: rhodes (ajr2283) – 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 ) − g ( x ) , B. lim x → 1 g ( x ) G ( x ) , C. lim x → 1 G ( x ) f ( x ) , are indeterminate forms? 1. B only 2. A and C only 3. C only 4. A and B only 5. A only 6. all of them correct 7. none of them 8. B and C only Explanation: A. Since lim x → 1 G ( x ) − g ( x ) = ∞− ∞ , this limit is an indeterminate form. B. Since lim x → 1 g ( x ) G ( x ) = ∞ ∞ , this limit is an indeterminate form. C. Since lim x → 1 G ( x ) f ( x ) = ∞ , this limit is an indeterminate form. 002 0.0 points Determine lim x → 1 e 7 x − e 7 ln(6 x − 5) . 1. limit = e 7 2. limit = 1 e 7 3. limit does not exist 4. limit = e 7 6 5. limit = 7 6 e 7 correct 6. limit = 7 6 Explanation: Since e 7 x − e 7 ln(6 x − 41) = 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 ) = 7 e 7 x , g ′ ( x ) = 6 6 x − 5 , so lim x → 1 f ′ ( x ) = 7 e 7 , lim x → 1 g ′ ( x ) = 6 . Consequently, lim x → 1 e 7 x − e 7 ln(6 x − 5) = 7 6 e 7 . rhodes (ajr2283) – ReviewExam01 – Gilbert – (56380) 2 003 0.0 points Find the value of lim x → 1 − cos 3 x 5 sin 2 2 x . 1. limit = 1 4 2. limit = 11 40 3. limit = 3 10 4. limit = 9 40 correct 5. limit does not exist Explanation: Set f ( x ) = 1 − cos 3 x, g ( x ) = 5 sin 2 2 x . Then f, g are differentiable functions such that lim x → f ( x ) = 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 ) = lim x → 3 sin3 x 20 sin2 x cos 2 x . To compute this last limit we can either apply L’Hospital’s Rule again or use the fact that lim x → sin3 x x = 3 , lim x → sin 2 x x = 2 . Consequently, lim x → 1 − cos 3 x 5 sin 2 2 x = 9 40 . 004 0.0 points Determine if the improper integral I = integraldisplay 1 5 ln2 x dx converges, and if it does, find its value. 1. I = 5 ln2 − 2 2. I = 2 ln5 + 5 3. I = 2(ln 5 + 1) 4. I = 5(ln 2 − 1) correct 5. I = 2(ln 5 − 1) 6. I does not converge 7. I = 5(ln 2 + 1) Explanation: Since ln 2 x → −∞ as x → 0 through pos- itive values of x , the graph of 5 ln2 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 5 ln2 x dx and check if the limit exists.and check if the limit exists....
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Review Exam 1 - rhodes(ajr2283 – ReviewExam01 – Gilbert...

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