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Unformatted text preview: padilla (tp5647) – HW01 – Gilbert – (56650) 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 →∞ 7 x 2 − 5 x + 6 5 + 7 x − 3 x 2 . 1. none of the other answers 2. limit = ∞ 3. limit = 0 4. limit = − 7 3 correct 5. limit = 7 6 Explanation: Dividing the numerator and denominator by x 2 we see that 7 x 2 − 5 x + 6 5 + 7 x − 3 x 2 = 7 − 5 x + 6 x 2 5 x 2 + 7 x − 3 . On the other hand, lim x →∞ 1 x = lim x →∞ 1 x 2 = 0 . By Properties of limits, therefore, the limit = − 7 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 3 and Q ( x ) has degree 5. 1. limit = ∞ 2. not enough information given 3. limit = 0 correct 4. limit = 3 5. limit = −∞ 6. limit = 5 Explanation: Since P has degree 3 and Q has degree 5, there exist a, b negationslash = 0 such that P ( x ) = ax 3 + R ( x ) , Q ( x ) = bx 5 + S ( x ) where R ( x ) , S ( x ) are polynomials such that deg( R ) < 3 , deg( S ) < 5. Thus P ( x ) Q ( x ) = ax 3 + R ( x ) bx 5 + S ( x ) = a x 2 + R ( x ) x 5 b + S ( x ) x 5 . On the other hand, lim x →∞ 1 x 2 = lim x →∞ R ( x ) x 5 = lim x →∞ S ( x ) x 5 = 0 since deg( R ) , deg( S ) < 5. Consequently, by Properties of Limits, lim x →∞ P ( x ) Q ( x ) = 0 . 003 10.0 points Determine if lim x →∞ braceleftBig ln(3 + 4 x ) − ln(8 + 5 x ) bracerightBig exists, and if it does find its value. 1. limit = ln 8 3 2. limit does not exist 3. limit = ln 4 5 correct padilla (tp5647) – HW01 – Gilbert – (56650) 2 4. limit = ln 3 8 5. limit = ln 5 4 Explanation: By properties of logs, ln(3 + 4 x ) − ln(8 + 5 x ) = ln parenleftbigg 3 + 4 x 8 + 5 x parenrightbigg = ln parenleftbigg 3 /x + 4 8 /x + 5 parenrightbigg . But lim x →∞ 3 /x + 4 8 /x + 5 = 4 5 . Consequently, the limit exists and limit = ln 4 5 . 004 10.0 points Find the value of lim x →∞ parenleftbigg 3 e x − 4 e − x 2 e x + 5 e − x parenrightbigg . 1. limit = − 3 2 2. limit = − 2 3 3. limit = 2 3 4. limit = 1 7 5. limit = − 1 7 6. limit = 3 2 correct Explanation: After division we see that 3 e x − 4 e − x 2 e x + 5 e − x = 3 − 4 e − 2 x 2 + 5 e − 2 x ....
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This note was uploaded on 05/03/2009 for the course M 408 L taught by Professor Gilbert during the Spring '09 term at University of TexasTyler.
 Spring '09
 GILBERT
 Calculus

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