Homework #3 - jiwani (amj566) Homework03 Fouli (58395) 1...

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Unformatted text preview: jiwani (amj566) Homework03 Fouli (58395) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Suppose lim x 5 f ( x ) = 4 . Which of these statements are true without further restrictions on f ? A. Range of f contains 4 . B. f is defined on ( a, b ) for some a < 5 < b . C. As x approaches 5 , f ( x ) approaches 4 . 1. B and C only 2. A and B only 3. B only 4. None of them 5. C only correct 6. All of them 7. A and C only 8. A only Explanation: A. Not True: f ( x ) need only approach 4. B. Not true: f ( x ) need not be defined at x = 5. C. True: definition of limit keywords: True/False, definition limit, limit, domain, range, 002 10.0 points Below is the graph of a function f . 2 4 6 2 4 6 2 4 6 8 2 4 Use the graph to determine lim x 2- f ( x ) . 1. limit does not exist 2. limit = 0 3. limit = 4 4. limit = 4 correct 5. limit = 2 Explanation: As the graph shows, lim x 2- f ( x ) = 4 . 003 10.0 points A function f is defined piecewise for all x negationslash = 0 by f ( x ) = 4 + 1 2 x, x < 2 , 3 2 x, < | x | 2 , 3 + 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. jiwani (amj566) Homework03 Fouli (58395) 2 1. ( , 2) ( 2 , 2) (2 , ) 2. ( , 2) (2 , ) 3. ( , 0) (0 , 2) (2 , ) 4. ( , 2) ( 2 , ) correct 5. ( , 2) ( 2 , 0) (0 , 2) (2 , ) 6. ( , 2) ( 2 , 0) (0 , ) 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 , ) . 004 10.0 points If f oscillates faster and faster when x ap- proaches 0 as indicated by its graph determine which, if any, of L 1 : lim x 0+ f ( x ) , L 2 : lim x - f ( x ) exist. 1. L 1 doesnt exist, but L 2 does correct 2. neither L 1 nor L 2 exists 3. L 1 exists, but L 2 doesnt 4. both L 1 and L 2 exist Explanation: For x > 0 the graph of f oscillates but the oscillations do not get smaller and smaller as x approaches 0; so L 1 does not exist. But for x < 0, the graph oscillates and the oscillations get smaller and smaller as x approaches 0; in fact, the oscillation goes to 0 as x approaches 0, so L 2 exists. Consequently, L 1 does not exist, but L 2 does . 005 10.0 points Suppose that f ( x ) is defined for all x in U = (2 , 3) (3 , 4) and that lim x 3 f ( x ) = L....
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Homework #3 - jiwani (amj566) Homework03 Fouli (58395) 1...

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