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# Exam1 - Betancourt Daniel Exam 1 Due 11:00 pm Inst Diane...

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Betancourt, Daniel – Exam 1 – Due: Feb 20 2007, 11:00 pm – Inst: Diane Radin 1 This print-out should have 20 questions. Multiple-choice 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 ) -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 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 0 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 > 0 for all small x 6 = 0, both positive and negative, and lim x 0 x sin x = 0 , we see that lim x 0 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

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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 . Alternatively, if we recognize that lim x 2 f ( x ) - f (2) x - 2 = f 0 (2) = (6 x - 5) | x = 2 ,
Betancourt, Daniel – Exam 1 – Due: Feb 20 2007, 11:00 pm – Inst: Diane Radin 3 we see again that lim x 2 f ( x ) - f (2) x - 2 = 7 .

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Exam1 - Betancourt Daniel Exam 1 Due 11:00 pm Inst Diane...

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