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Unformatted text preview: Rehman (aar638) – HW01 – sachse – (56620) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine lim x →∞ x 2 − 8 x + 7 6 + 5 x − 3 x 2 . 1. limit = ∞ 2. limit = − 1 3 correct 3. none of the other answers 4. limit = 0 5. limit = 1 6 Explanation: Dividing the numerator and denominator by x 2 we see that x 2 − 8 x + 7 6 + 5 x − 3 x 2 = 1 − 8 x + 7 x 2 6 x 2 + 5 x − 3 . On the other hand, lim x →∞ 1 x = lim x →∞ 1 x 2 = 0 . By Properties of limits, therefore, the limit = − 1 3 . 002 10.0 points Let P ( x ) and Q ( x ) be polynomials. Deter mine lim x →∞ P ( x ) Q ( x ) when P ( x ) has degree 2 and Q ( x ) has degree 4. 1. limit = −∞ 2. limit = ∞ 3. limit = 0 correct 4. limit = 4 5. limit = 2 6. not enough information given Explanation: Since P has degree 2 and Q has degree 4, there exist a, b negationslash = 0 such that P ( x ) = ax 2 + R ( x ) , Q ( x ) = bx 4 + S ( x ) where R ( x ) , S ( x ) are polynomials such that deg( R ) < 2 , deg( S ) < 4. Thus P ( x ) Q ( x ) = ax 2 + R ( x ) bx 4 + S ( x ) = a x 2 + R ( x ) x 4 b + S ( x ) x 4 . On the other hand, lim x →∞ 1 x 2 = lim x →∞ R ( x ) x 4 = lim x →∞ S ( x ) x 4 = 0 since deg( R ) , deg( S ) < 4. Consequently, by Properties of Limits, lim x →∞ P ( x ) Q ( x ) = 0 . 003 10.0 points Determine if lim x →∞ braceleftBig ln(7 + 4 x ) − ln(2 + 5 x ) bracerightBig exists, and if it does find its value. 1. limit = ln 4 5 correct 2. limit = ln 5 4 3. limit does not exist Rehman (aar638) – HW01 – sachse – (56620) 2 4. limit = ln 7 2 5. limit = ln 2 7 Explanation: By properties of logs, ln(7 + 4 x ) − ln(2 + 5 x ) = ln parenleftbigg 7 + 4 x 2 + 5 x parenrightbigg = ln parenleftbigg 7 /x + 4 2 /x + 5 parenrightbigg . But lim x →∞ 7 /x + 4 2 /x + 5 = 4 5 . Consequently, the limit exists and limit = ln 4 5 . 004 10.0 points Find the value of lim x →∞ parenleftbigg 4 e x + 5 e − x 2 e x − 3 e − x parenrightbigg . 1. limit = 1 5 2. limit = − 1 5 3. limit = 2 correct 4. limit = − 2 5. limit = 1 2 6. limit = − 1 2 Explanation: After division we see that 4 e x + 5 e − x 2 e x − 3 e − x = 4 + 5 e − 2 x 2 − 3 e − 2 x ....
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This note was uploaded on 09/24/2010 for the course M408L m408l taught by Professor Radin during the Spring '08 term at University of Texas Health Science Center at San Antonio.
 Spring '08
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