Review Assignment-solutions HW 7

Review Assignment-solutions HW 7 - handa (nh5757) Review...

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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 – fnd 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 :l i m x 0+ f ( x ) ,L 2 i m x 0 - 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: ±or x> 0thegrapho F 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 3 ( x - 3) 2 ,x 3 . ±ind 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 )
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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 0 x 2 cos ± 10 x ² exists, and if it does, Fnd its value. 1. limit = 10 2. limit = 0 correct 3. limit = 1 4. limit = 2 5. limit does not exist Explanation: Since | cos θ |≤ 1fora l l θ ,weseethat ³ ³ ³ cos ± 10 x ² ³ ³ ³ 1 , holds for all x ± =0.Butthen ³ ³ ³ x 2 cos ± 10 x ² ³ ³ ³ ≤| x 2 | , for all x ± =0;inotherwords ,ifweset g ( x )= -| x 2 | ,h ( x | x 2 | , then the inequalities g ( x ) x 2 cos ± 10 x ² h ( x ) hold for all x ± =0.But lim x 0 g ( x )= l im x 0 h ( x )=0 . Consequently, by the Squeeze Theorem, the given limit lim x 0 x 2 cos ± 10 x ² exists and limit = 0 . keywords: limit, trigonometric function, squeeze theorem 004 10.0 points Determine if lim x 0 1 5 x sin x exists, and if it does, Fnd its value. 1. limit = 1 5 2. limit = - 1 5 3. limit = -∞ 4. none of the other answers 5. limit = correct Explanation: Since 1 5 x sin x > 0 for all small x ± =0,bothpos itiveandnegative , while lim x 0 x sin x =0 , we see that lim x 0 1 5 x sin x = . 005 10.0 points Determine the value of lim x 3 f ( x ) when f satisFes the inequalities 4 f ( x ) x 2 - 6 x +13 on [ - 5 , 3) (3 , 5].
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Review Assignment-solutions HW 7 - handa (nh5757) Review...

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