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Unformatted text preview: silva (jrs4378) – Hw 03 – gualdani – (56455) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points (i) Determine the value of lim x → 1+ x − 2 x − 1 . 1. limit = ∞ 2. limit = − 2 3. limit = −∞ correct 4. none of the other answers 5. limit = 2 Explanation: For 1 < x < 2 we see that x − 2 x − 1 < . On the other hand, lim x → 1+ x − 1 = 0 . Thus, by properties of limits, lim x → 1+ x − 2 x − 1 = −∞ . 002 (part 2 of 3) 10.0 points (ii) Determine the value of lim x → 1 x − 2 x − 1 . 1. limit = ∞ correct 2. none of the other answers 3. limit = − 2 4. limit = −∞ 5. limit = 2 Explanation: For x < 1 < 2 we see that x − 2 x − 1 > . On the other hand, lim x → 1 x − 1 = 0 . Thus, by properties of limits, lim x → 1 x − 2 x − 1 = ∞ . 003 (part 3 of 3) 10.0 points (iii) Determine the value of lim x → 1 x − 2 x − 1 . 1. limit = −∞ 2. limit = ∞ 3. limit = − 2 4. none of the other answers correct 5. limit = 2 Explanation: If lim x → 1 x − 2 x − 1 exists, then lim x → 1+ x − 2 x − 1 = lim x → 1 x − 2 x − 1 . But as parts (i) and (ii) show, lim x → 1+ x − 2 x − 1 negationslash = lim x → 1 x − 2 x − 1 . Consequently, lim x → 1 x − 2 x − 1 does not exist . silva (jrs4378) – Hw 03 – gualdani – (56455) 2 004 10.0 points Below is the graph of a function f . 2 4 − 2 − 4 2 4 − 2 − 4 Use the graph to determine lim x → 3 f ( x ). 1. lim x → 3 f ( x ) = − 2 2. lim x → 3 f ( x ) = 1 3. lim x → 3 f ( x ) does not exist correct 4. lim x → 3 f ( x ) = − 1 5. lim x → 3 f ( x ) = 0 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 . 005 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 → 3 f ( x ) . 1. limit does not exist 2. limit = 5 correct 3. limit = − 2 4. limit = 3 2 5. limit = − 4 Explanation: As the graph shows, lim x → 3 f ( x ) = 5 . 006 10.0 points If f oscillates faster and faster when x ap proaches 0 as indicated by its graph silva (jrs4378) – Hw 03 – gualdani – (56455) 3 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. neither L 1 nor L 2 exists 4. L 1 doesn’t exist, but L 2 does correct 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...
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This note was uploaded on 08/29/2010 for the course M 46455 taught by Professor Gualdani during the Spring '10 term at University of Texas.
 Spring '10
 Gualdani

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