This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: jiwani (amj566) – Homework03 – Fouli – (58395) 1 This printout should have 24 questions. Multiplechoice 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 doesn’t exist, but L 2 does correct 2. neither L 1 nor L 2 exists 3. L 1 exists, but L 2 doesn’t 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....
View
Full Document
 Spring '11
 Jensen
 Limit, lim, Limit of a function

Click to edit the document details