Calculus 408K Exam 1

# Calculus 408K Exam 1 - Barkley, Thane Exam 1 Due: Oct 2...

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Unformatted text preview: Barkley, Thane Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Louiza Fouli 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 x 16 + 3 x- 4 . 1. limit = 2. limit = 3 8 3. limit = 4 3 4. limit = 8 3 correct 5. limit = 0 6. limit = 3 4 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 16 + 3 x = 4 . Consequently, by properties of limits, lim x 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 5 x f ( x ) 1 3 x 3 + 4 x + 2 3 on [0 , 1) (1 , 2]. 1. limit does not exist 2. limit = 4 3. limit = 5 correct 4. limit = 3 5. limit = 2 6. limit = 6 Explanation: Set g ( x ) = 5 x, h ( x ) = 1 3 x 3 + 4 x + 2 3 . Then, by properties of limits, lim x 1 g ( x ) = lim x 1 5 x = 5 , while lim x 1 h ( x ) = lim x 1 1 3 x 3 + 4 x + 2 3 = 1 3 + 4 + 2 3 = 5 . By the Squeeze Theorem, therefore, lim x 1 f ( x ) = 5 . Barkley, Thane Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Louiza Fouli 2 To see why the Squeeze theorem applies, its a good idea to draw the graphs of g and h using, say, a graphing calculator. They look like g : h : (1 , 5) (not drawn to scale), so the graphs of g and h touch at the point (1 , 5) while the graph of f is sandwiched between these two graphs. Thus again we see that lim x 1 f ( x ) = 5 . keywords: limit, squeeze theorem 003 (part 1 of 1) 10 points Let f be the function defined by f ( x ) = x + ( x- 2 + | x- 2 | ) 2 . Determine if lim h f (3 + h )- f (3) h exists, and if it does, find its value. 1. limit doesnt exist 2. limit = 6 3. limit = 10 4. limit = 8 5. limit = 9 correct 6. limit = 7 Explanation: Since | v | = v, v 0,- v, v < 0, we see that x- 2 + | x- 2 | = 2( x- 2) , x 2, , x < 2. Thus f ( x ) = 1 x, x < 2, 1 x + 4( x- 2) 2 , x 2. In particular, therefore, lim h f (3 + h )- f (3) h = d dx 1 x + 4( x- 2) 2 fl fl fl x = 3 = 1 + 8( x- 2) fl fl fl x = 3 because 3 , 3 + h > 2 for all small h . Conse- quently, limit = 9 . keywords: limit, Newtonian quotient absolute value function 004 (part 1 of 1) 10 points Determine if the limit lim x sin 2 x 6 x exists, and if it does, find its value. 1. limit = 3 2. limit = 2 3. limit = 6 Barkley, Thane Exam 1 Due: Oct 2 2007, 11:00 pm Inst: Louiza Fouli 3 4. limit = 1 3 correct 5. limit doesnt exist Explanation: Using the known limit: lim x sin ax x = a , we see that lim x sin 2 x 6 x = 1 3 ....
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## This test prep was uploaded on 04/16/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.

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Calculus 408K Exam 1 - Barkley, Thane Exam 1 Due: Oct 2...

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