# CalcHW3Answers - Villarreal Natalie Homework 3 Due 3:00 am...

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Villarreal, Natalie – Homework 3 – Due: Sep 11 2007, 3:00 am – Inst: Louiza Fouli 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 3) 10 points (i) Determine the value o± lim x 2+ x - 6 x - 2 . 1. limit = - 3 2. limit = 3 3. limit = -∞ correct 4. limit = 5. none o± the other answers Explanation: For 2 < x < 6 we see that x - 6 x - 2 < 0 . On the other hand, lim x 2+ x - 2 = 0 . Thus, by properties o± limits, lim x 2+ x - 6 x - 2 = -∞ . 002 (part 2 o± 3) 10 points (ii) Determine the value o± lim x 2 - x - 6 x - 2 . 1. limit = 3 2. limit = -∞ 3. limit = correct 4. none o± the other answers 5. limit = - 3 Explanation: For x < 2 < 6 we see that x - 6 x - 2 > 0 . On the other hand, lim x 2 - x - 2 = 0 . Thus, by properties o± limits, lim x 2 - x - 6 x - 2 = . 003 (part 3 o± 3) 10 points (iii) Determine the value o± lim x 2 x - 6 x - 2 . 1. limit = - 3 2. none o± the other answers correct 3. limit = 3 4. limit = -∞ 5. limit = Explanation: lim x 2 x - 6 x - 2 exists, then lim x 2+ x - 6 x - 2 = lim x 2 - x - 6 x - 2 . But as parts (i) and (ii) show, lim x 2+ x - 6 x - 2 6 = lim x 2 - x - 6 x - 2 . Consequently, lim x 2 x - 6 x - 2 does not exist .

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Villarreal, Natalie – Homework 3 – Due: Sep 11 2007, 3:00 am – Inst: Louiza Fouli 2 keywords: limit, left hand limit, right hand limit, rational function, 004 (part 1 of 1) 10 points Suppose that f ( x ) is de±ned for all x in U = (1 , 2) (2 , 3) and that lim x 2 f ( x ) = L. Which of the following statements is then true? I) If L > 0, then f ( x ) > 0 on U . II) If f ( x ) > 0 on U , then L 0. III) If L = 0, then f ( x ) = 0 on U . 1. II only correct 2. None of these 3. each of I, II, III 4. I, III only 5. I, II only 6. II, III only Explanation: I) False: consider the function f ( x ) = 1 - 2 | x - 2 | . Its graph is 2 4 6 so lim x 2 f ( x ) = 1 . But on (1 , 3 2 ) and on ( 5 2 , 3) we see that f ( x ) < 0. II) True: if f ( x ) > 0 on U , then on U the graph of f always lies above the x -axis. So as x approaches 2, the point ( x, f ( x )) on the graph approaches the point (2 , L ). Thus L 0; notice that L = 0 can occur as the graph 2 4 6 2 of f ( x ) = | x - 2 | shows.
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CalcHW3Answers - Villarreal Natalie Homework 3 Due 3:00 am...

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