calc3 - pokharel (yp624) – HW03 – Radin – (56520) 1...

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Unformatted text preview: pokharel (yp624) – HW03 – Radin – (56520) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 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. limit = 0 2. does not exist 3. limit = 2 4. limit = − 1 correct 5. limit = 1 Explanation: From the graph it is clear that the limit lim x → 3 − f ( x ) = − 1 , from the left and the limit lim x → 3+ f ( x ) = − 1 , from the right exist and coincide in value. Thus the two-sided lim x → 3 f ( x ) = − 1 . 002 10.0 points Below is the graph of a function y = f ( x ) 2 4 6 − 2 − 4 − 6 2 4 6 8 − 2 − 4 Use the graph to determine lim x → 1+ f ( x ) . 1. limit = − 2 correct 2. limit does not exist 3. limit = 1 2 4. limit = 3 5. limit = − 4 Explanation: From the graph we see that lim x → 1+ f ( x ) = − 2 . 003 10.0 points If f oscillates faster and faster when x ap- proaches 0 as indicated by its graph pokharel (yp624) – HW03 – Radin – (56520) 2 determine which, if any, of L 1 : lim x → 0+ f ( x ) , L 2 : lim x → − f ( x ) exist. 1. neither L 1 nor L 2 exists 2. L 1 exists, but L 2 doesn’t correct 3. both L 1 and L 2 exist 4. L 1 doesn’t exist, but L 2 does Explanation: For x > 0 the graph of f oscillates but the oscillations get smaller and smaller as x approaches 0; in fact, the oscillation goes to 0 as x approaches 0, so L 1 exists. But for x < 0, the graph oscillates but the oscillations do not get smaller and smaller, so L 2 does not exist. Consequently, L 1 exists, but L 2 doesn’t . 004 10.0 points Consider the function f ( x ) = 3 − x, x < − 1 x, − 1 ≤ x < 2 ( x − 3) 2 , x ≥ 2 . Find all the values of a for which the limit lim x → a f ( x ) exists, expressing your answer in interval no- tation. 1. ( −∞ , 2) ∪ (2 , ∞ ) 2. ( −∞ , − 1] ∪ [2 , ∞ ) 3. ( −∞ , − 1) ∪ ( − 1 , 2) ∪ (2 , ∞ ) correct 4. ( −∞ , − 1) ∪ ( − 1 , ∞ ) 5. ( −∞ , ∞ ) Explanation: The graph of f is a straight line on ( −∞ , − 1), so lim x → a f ( x ) exists (and = f ( a )) for all a in ( −∞ , − 1). Similarly, the graph of f on ( − 1 , 2) is a straight line, so lim x → a f ( x ) exists (and = f ( a )) for all a in ( − 1 , 2). On (2 , ∞ ), however, the graph of f is a parabola, so lim x → a f ( x ) still exists (and = f ( a )) for all a in (2 , ∞ ). On the other hand, lim x →− 1 − f ( x ) = 4 , lim x →− 1+ f ( x ) = − 1 , while lim x → 2 − f ( x ) = 2 , lim x → 2+ f ( x ) = 1 . Thus neither of the limits lim x →− 1 f ( x ) , lim x → 2 f ( x ) exists. Consequently, the limit exists only for values of a in ( −∞ , − 1) ∪ ( − 1 , 2) ∪ (2 , ∞ ) ....
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This note was uploaded on 06/05/2010 for the course PHYS 92515 taught by Professor Tsoi during the Spring '10 term at University of Texas.

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calc3 - pokharel (yp624) – HW03 – Radin – (56520) 1...

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