Solutions HW 3

# Solutions HW 3 - mao(tm23477 – HW3 – gualdani –(57180...

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Unformatted text preview: mao (tm23477) – HW3 – gualdani – (57180) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Suppose that f ( x ) is defined for all x in U = (4 , 5) ∪ (5 , 6) and that lim x → 5 f ( x ) = L. Which of the following statements are then true? A. if L > , then f ( x ) > on U ; B. if f ( x ) > on U , then L ≥ 0; C. if L = 0 , then f ( x ) = 0 on U . 1. A, B only 2. B , C only 3. A, C only 4. none of A, B , C 5. B only correct 6. each of A, B , C Explanation: A. False: consider the function f ( x ) = 1 − 2 | x − 5 | . Its graph is 2 4 6 so lim x → 5 f ( x ) = 1 . But on (4 , 9 2 ) and on ( 11 2 , 6) we see that f ( x ) < 0. B. True: if f ( x ) > 0 on U , then on U the graph of f always lies above the x-axis. So as x approaches 5, the point ( x, f ( x )) on the graph approaches the point (5 , L ). Thus L ≥ 0; notice that L = 0 can occur as the graph 2 4 6 2 of f ( x ) = | x − 5 | shows. C. False: consider the function f ( x ) = | x − 5 | . Its graph 2 4 6 2 shows that f ( x ) > 0 for all x negationslash = 5, but lim x → 5 | x − 5 | = 0 . keywords: limit, definition of limit, properties of limits, T/F, True/False 002 (part 1 of 2) 10.0 points (i) Which of the following statements are true for all values of c ? I. lim x → c f ( x ) = 0 = ⇒ lim x → c | f ( x ) | = 0 . II. lim x → c | f ( x ) | = 0 = ⇒ lim x → c f ( x ) = 0 . 1. both I and II correct 2. neither I nor II 3. II only mao (tm23477) – HW3 – gualdani – (57180) 2 4. I only Explanation: 003 (part 2 of 2) 10.0 points (ii) Which of the following statements are true for all c and all L ? I. lim x → c f ( x ) = L = ⇒ lim x → c | f ( x ) | = | L | . II. lim x → c | f ( x ) | = | L | = ⇒ lim x → c f ( x ) = L. 1. II only 2. I only correct 3. neither I nor II 4. both I and II Explanation: 004 10.0 points A function f is defined piecewise for all x negationslash = 0 by f ( x ) = 3 + 1 2 x, x < − 2 , 5 2 x, < | x | ≤ 2 , 5 + x − 1 2 x 2 , x > 2 . By first drawing the graph of f , determine all the values of a at which lim x → a f ( x ) exists, expressing your answer in interval no- tation. 1. ( −∞ , − 2) ∪ ( − 2 , 0) ∪ (0 , 2) ∪ (2 , ∞ ) 2. ( −∞ , − 2) ∪ ( − 2 , 2) ∪ (2 , ∞ ) 3. ( −∞ , 0) ∪ (0 , 2) ∪ (2 , ∞ ) 4. ( −∞ , 2) ∪ (2 , ∞ ) 5. ( −∞ , − 2) ∪ ( − 2 , 0) ∪ (0 , ∞ ) 6. ( −∞ , − 2) ∪ ( − 2 , ∞ ) correct 7. ( −∞ , 0) ∪ (0 , ∞ ) Explanation: The graph of f is 2 4 − 2 − 4 2 4 − 2 − 4 and inspection shows that lim x → a f ( x ) will exist only for a in ( −∞ , − 2) ∪ ( − 2 ,...
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## This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.

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Solutions HW 3 - mao(tm23477 – HW3 – gualdani –(57180...

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