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# Exam1_2 - Taylor Douglas Exam 1 Due Oct 2 2007 11:00 pm...

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Taylor, Douglas – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the value of lim x 0 x 16 + 3 x - 4 . 1. limit = 3 4 2. limit = 8 3 correct 3. limit = 4 3 4. limit = 3 8 5. limit = 6. limit = 0 Explanation: After rationalization, 16 + 3 x - 4 = (16 + 3 x ) - 16 16 + 3 x + 4 . Thus f ( x ) = x 16 + 3 x - 4 = x ( 16 + 3 x + 4 ) 3 x , from which it follows that f ( x ) = 16 + 3 x + 4 3 for x 6 = 0. Now lim x 0 16 + 3 x = 4 . Consequently, by properties of limits, lim x 0 f ( x ) = 8 3 . keywords: limit, evaluate limit analytically, rationalize denominator, 002 (part 1 of 1) 10 points Determine the value of lim x 1 f ( x ) when f satisfies the inequalities 6 x f ( x ) 1 3 x 3 + 5 x + 2 3 on [0 , 1) (1 , 2]. 1. limit = 4 2. limit = 3 3. limit = 6 correct 4. limit does not exist 5. limit = 5 6. limit = 7 Explanation: Set g ( x ) = 6 x, h ( x ) = 1 3 x 3 + 5 x + 2 3 . Then, by properties of limits, lim x 1 g ( x ) = lim x 1 6 x = 6 , while lim x 1 h ( x ) = lim x 1 1 3 x 3 + 5 x + 2 3 · = 1 3 + 5 + 2 3 = 6 . By the Squeeze Theorem, therefore, lim x 1 f ( x ) = 6 .

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Taylor, Douglas – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert 2 To see why the Squeeze theorem applies, it’s a good idea to draw the graphs of g and h using, say, a graphing calculator. They look like g : h : (1 , 6) (not drawn to scale), so the graphs of g and h ‘touch’ at the point (1 , 6) while the graph of f is ‘sandwiched’ between these two graphs. Thus again we see that lim x 1 f ( x ) = 6 . keywords: limit, squeeze theorem 003 (part 1 of 1) 10 points Let f be the function defined by f ( x ) = 3 x + ( x - 2 + | x - 2 | ) 2 . Determine if lim h 0 f (4 + h ) - f (4) h exists, and if it does, find its value. 1. limit = 15 2. limit doesn’t exist 3. limit = 16 4. limit = 18 5. limit = 17 6. limit = 19 correct Explanation: Since | v | = v, v 0, - v, v < 0, we see that x - 2 + | x - 2 | = 2( x - 2) , x 2, 0 , x < 2. Thus f ( x ) = 3 x, x < 2, 3 x + 4( x - 2) 2 , x 2. In particular, therefore, lim h 0 f (4 + h ) - f (4) h = d dx 3 x + 4( x - 2) 2 ·fl fl fl x = 4 = 3 + 8( x - 2) ·fl fl fl x = 4 because 4 , 4 + h > 2 for all small h . Conse- quently, limit = 19 . keywords: limit, Newtonian quotient absolute value function 004 (part 1 of 1) 10 points Determine if the limit lim x 0 sin 5 x 6 x exists, and if it does, find its value. 1. limit = 6 2. limit = 6 5 3. limit = 5 6 correct 4. limit = 5
Taylor, Douglas – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert 3 5. limit doesn’t exist Explanation: Using the known limit: lim x 0 sin ax x = a , we see that lim x 0 sin 5 x 6 x = 5 6 . keywords: limit, trigonometric function 005 (part 1 of 1) 10 points Determine if lim x 1 x 2 - 3 x + 2 x 2 + 2 x - 3 exists, and if it does, find its value. 1. limit = 1 4 2. limit = 1 5 3. limit = - 1 2 4. limit = - 1 5 5. limit does not exist 6. limit = - 1 4 correct Explanation: After factoring the numerator and denomi- nator we see that x 2 - 3 x + 2 x 2 + 2 x - 3 = ( x - 1)( x - 2) ( x - 1)( x + 3) = x - 2 x + 3 provided x 6 = 1. On the other hand, by properties of limits, lim x 1 x - 2 x + 3 = - 1 4 .

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Exam1_2 - Taylor Douglas Exam 1 Due Oct 2 2007 11:00 pm...

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