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Unformatted text preview: griffin (ang684) 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 2 x 2 6 x + 4 4 + 6 x 4 x 2 . 1. limit = 0 2. limit = 1 2 correct 3. none of the other answers 4. limit = 1 4 5. limit = Explanation: Dividing the numerator and denominator by x 2 we see that 2 x 2 6 x + 4 4 + 6 x 4 x 2 = 2 6 x + 4 x 2 4 x 2 + 6 x 4 . On the other hand, lim x 1 x = lim x 1 x 2 = 0 . By Properties of limits, therefore, the limit = 1 2 . 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 7. 1. limit = 2 2. not enough information given 3. limit = 4. limit = 5. limit = 7 6. limit = 0 correct Explanation: Since P has degree 2 and Q has degree 7, there exist a, b negationslash = 0 such that P ( x ) = ax 2 + R ( x ) , Q ( x ) = bx 7 + S ( x ) where R ( x ) , S ( x ) are polynomials such that deg( R ) < 2 , deg( S ) < 7. Thus P ( x ) Q ( x ) = ax 2 + R ( x ) bx 7 + S ( x ) = a x 5 + R ( x ) x 7 b + S ( x ) x 7 . On the other hand, lim x 1 x 5 = lim x R ( x ) x 7 = lim x S ( x ) x 7 = 0 since deg( R ) , deg( S ) < 7. Consequently, by Properties of Limits, lim x P ( x ) Q ( x ) = 0 . 003 10.0 points Determine if lim x braceleftBig ln(1 + 8 x ) ln(3 + x ) bracerightBig exists, and if it does find its value. 1. limit = ln 3 2. limit = ln 1 8 3. limit = ln 1 3 4. limit does not exist griffin (ang684) HW01 Gilbert (56650) 2 5. limit = ln 8 correct Explanation: By properties of logs, ln(1 + 8 x ) ln(3 + x ) = ln parenleftbigg 1 + 8 x 3 + x parenrightbigg = ln parenleftbigg 1 /x + 8 3 /x + 1 parenrightbigg . But lim x 1 /x + 8 3 /x + 1 = 8 . Consequently, the limit exists and limit = ln 8 . 004 10.0 points Find the value of lim x parenleftbigg e x 5 e x 5 e x + 4 e x parenrightbigg . 1. limit = 5 2. limit = 4 9 3. limit = 1 5 correct 4. limit = 1 5 5. limit = 4 9 6. limit = 5 Explanation: After division we see that e x 5 e x 5 e x + 4 e x = 1 5 e 2 x 5 + 4 e 2 x ....
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This note was uploaded on 09/09/2009 for the course M 408L taught by Professor Gilbert during the Spring '09 term at University of TexasTyler.
 Spring '09
 GILBERT

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