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Unformatted text preview: Betancourt, Daniel Exam 1 Due: Feb 20 2007, 11:00 pm Inst: Diane Radin 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Below is the graph of a function y = f ( x ) 2 4 6 2 4 6 2 4 6 8 2 4 Use the graph to determine the right hand limit lim x 1+ f ( x ) . 1. the limit does not exist 2. lim x 1+ f ( x ) = 1 3. lim x 1+ f ( x ) = 1 2 4. lim x 1+ f ( x ) = 0 correct 5. lim x 1+ f ( x ) = 4 Explanation: From the graph lim x 1+ f ( x ) = 0. keywords: graph, limit right hand limit 002 (part 1 of 1) 10 points Determine lim x 1 8 x sin x . 1. limit = 1 8 2. limit = correct 3. limit = 1 8 4. none of these 5. limit = Explanation: Since 1 8 x sin x > for all small x 6 = 0, both positive and negative, and lim x x sin x = 0 , we see that lim x 1 8 x sin x = . keywords: limit, trigonometric function 003 (part 1 of 1) 10 points Determine lim x 1 n 1 x 1 1 x 2 x o . 1. limit does not exist 2. limit = 1 correct 3. limit = 1 3 4. limit = 1 3 5. limit = 1 2 Betancourt, Daniel Exam 1 Due: Feb 20 2007, 11:00 pm Inst: Diane Radin 2 6. limit = 1 2 7. limit = 1 Explanation: After simplification we see that 1 x 1 1 x 2 x = x 1 x ( x 1) = 1 x , for all x 6 = 1. Thus limit = lim x 1 1 x = 1 . keywords: analytic limit, difference rational functions, limit, common denominators 004 (part 1 of 1) 10 points Evaluate lim x  1 x + 1 x 2 2 x 3 . 1. limit does not exist 2. limit = 1 4 3. limit = 1 2 4. limit = 2 5. limit = 1 2 6. limit = 1 4 correct Explanation: Since x 2 2 x 3 = ( x 3)( x + 1) , the expression above can be rewritten as x + 1 x 2 2 x 3 = 1 x 3 for x 6 = 1. Thus lim x  1 x + 1 x 2 2 x 3 = 1 4 . keywords: limit, rational function 005 (part 1 of 1) 10 points Determine lim x 2 f ( x ) f (2) x 2 when f ( x ) = 3 x 2 5 x + 2 . 1. limit = 5 2. limit = 6 3. limit = 4 4. limit does not exist 5. limit = 7 correct 6. limit = 3 Explanation: Since f ( x ) f (2) = 3 x 2 5 x + 2  4 = 3 x 2 5 x 2 = (3 x + 1)( x 2) , we see that f ( x ) f (2) x 2 = 3 x + 1 , for x 6 = 2. Consequently, lim x 2 f ( x ) f (2) x 2 = 7 ....
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This note was uploaded on 02/04/2012 for the course MATH 408K taught by Professor Gualdani during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Gualdani

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