Review Assignment-solutions

Review Assignment-solutions - handa(nh5757 – Review...

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Unformatted text preview: handa (nh5757) – Review Assignment – bormashenko – (54880) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 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. neither L 1 nor L 2 exists 2. both L 1 and L 2 exist 3. L 1 exists, but L 2 doesn’t 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 L 2 does . 002 10.0 points Consider the function f ( x ) = 3 − 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. 1. ( −∞ , − 1] ∪ [3 , ∞ ) 2. ( −∞ , − 1) ∪ ( − 1 , 3) ∪ (3 , ∞ ) correct 3. ( −∞ , 3) ∪ (3 , ∞ ) 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 , 3) is a straight line, so lim x → a f ( x ) exists (and = f ( a )) for all a in ( − 1 , 3). On (3 , ∞ ), however, the graph of f is a parabola, so lim x → a f ( x ) still exists (and = f ( a )) for all a in (3 , ∞ ). On the other hand, lim x →− 1 − f ( x ) = 4 , lim x →− 1+ f ( x ) = − 1 , while lim x → 3 − f ( x ) = 3 , lim x → 3+ f ( x ) = 0 . Thus neither of the limits lim x →− 1 f ( x ) , lim x → 3 f ( x ) handa (nh5757) – Review Assignment – bormashenko – (54880) 2 exists. Consequently, the limit exists only for values of a in ( −∞ , − 1) ∪ ( − 1 , 3) ∪ (3 , ∞ ) . 003 10.0 points Determine if lim x → x 2 cos parenleftBig 10 x parenrightBig exists, and if it does, find its value. 1. limit = 10 2. limit = 0 correct 3. limit = 1 4. limit = 2 5. limit does not exist Explanation: Since | cos θ | ≤ 1 for all θ , we see that vextendsingle vextendsingle vextendsingle cos parenleftBig 10 x parenrightBigvextendsingle vextendsingle vextendsingle ≤ 1 , holds for all x negationslash = 0. But then vextendsingle vextendsingle vextendsingle x 2 cos parenleftBig 10 x parenrightBigvextendsingle vextendsingle vextendsingle ≤ | x 2 | , for all x negationslash = 0; in other words, if we set g ( x ) = −| x 2 | , h ( x ) = | x 2 | , then the inequalities g ( x ) ≤ x 2 cos parenleftBig 10 x parenrightBig ≤ h ( x ) hold for all x negationslash = 0. But lim x → g ( x ) = lim x → h ( x ) = 0 ....
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This note was uploaded on 11/21/2011 for the course M 408N taught by Professor Gualdini during the Spring '10 term at University of Texas.

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Review Assignment-solutions - handa(nh5757 – Review...

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