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Unformatted text preview: Version 028 – Homework 03 – Helleloid – (58250) 1 This printout should have 23 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. f is defined on ( a, 5) ∪ (5 , b ) for some a < 5 < b . B. As x approaches 5 , f ( x ) approaches 4 . C. Range of f contains 4 . 1. B and C only 2. A only 3. A and B only correct 4. None of them 5. C only 6. All of them 7. A and C only 8. B only Explanation: A. True: f ( x ) needs only be defined near x = 5. B. True: definition of limit C. Not True: f ( x ) needs only AP PROACH 4. keywords: Stewart5e, True/False, definition limit limit 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. lim x →  2 f ( x ) = 7 2. lim x →  2 f ( x ) = 12 3. lim x →  2 f ( x ) does not exist 4. lim x →  2 f ( x ) = 8 5. lim x →  2 f ( x ) = 4 correct Explanation: From the graph it is clear the f has both a left hand limit and a right hand limit at x = − 2; in addition, these limits coincide. Thus lim x → 2 f ( x ) = 4. 003 10.0 points Below is the graph of a function f . Version 028 – Homework 03 – Helleloid – (58250) 2 2 4 − 2 − 4 2 4 − 2 − 4 Use the graph to determine lim x → 3 f ( x ). 1. lim x → 3 f ( x ) = 1 2. lim x → 3 f ( x ) = − 2 3. lim x → 3 f ( x ) = − 1 4. lim x → 3 f ( x ) = 0 5. lim x → 3 f ( x ) does not exist correct Explanation: From the graph it is clear that f has a left hand limit at x = 3 which is equal to − 2; and a right hand limit which is equal to 0. Since the two numbers do not coincide, the limit lim x → 3 f ( x ) does not exist . 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. both L 1 and L 2 exist 2. L 1 exists, but L 2 doesn’t 3. L 1 doesn’t exist, but L 2 does correct 4. neither L 1 nor L 2 exists 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 Consider the function f ( x ) = 2 − x, x < − 1 x, − 1 ≤ x < 3 ( x − 3) 2 , x ≥ 3 . Find all the values of a for which the limit lim x → a f ( x ) exists, expressing your answer in interval no tation....
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This note was uploaded on 12/04/2009 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz
 Calculus

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