Exam 1 - Garcia Ilse – Exam 1 – Due Oct 2 2007 11:00 pm...

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Unformatted text preview: Garcia, Ilse – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: Fonken 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 . Garcia, Ilse – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: Fonken 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 , 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 doesn’t 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 Garcia, Ilse – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: Fonken 3 4. limit = 1 3 correct 5. limit doesn’t 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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Exam 1 - Garcia Ilse – Exam 1 – Due Oct 2 2007 11:00 pm...

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