K Final Exam-solutions

K Final Exam-solutions - Version 044 K Final Exam...

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Version 044 – K Final Exam – ben-zvi – (55600) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Below is the graph of a function f . 2 4 6 2 4 6 2 4 6 2 4 6 Use this graph to determine the value of lim x 1 f ( x ) . 1. limit = 4 2. limit = 3 3. limit = 7 4. limit does not exist 5. limit = 4 correct Explanation: From the graph it is clear that lim x 1 f ( x ) = 4 . 002 10.0points Find the value of lim x 5 2 x 10 x 5 if the limit exists. 1. limit = 4 5 correct 2. limit = 2 5 3. limit = 6 5 4. limit = 3 5 5. limit = 20 6. limit does not exist Explanation: Since x 5 = ( x + 5)( x 5) , we can rewrite the given expression as 2( x + 5)( x 5) x 5 = 2( x + 5) for x negationslash = 5. Thus lim x 5 2 x 10 x 5 = 4 5 . 003 10.0points Find the value of lim x 0 1 cos 3 x 3 sin 2 2 x . 1. limit = 11 24 2. limit = 1 2 3. limit = 5 12 4. limit = 3 8 correct 5. limit does not exist Explanation:
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Version 044 – K Final Exam – ben-zvi – (55600) 2 Set f ( x ) = 1 cos 3 x, g ( x ) = 3 sin 2 2 x . Then f, g are differentiable functions such that lim x 0 f ( x ) = lim x 0 g ( x ) = 0 . Thus L’Hospital’s Rule can be applied: lim x 0 f ( x ) g ( x ) = lim x 0 f ( x ) g ( x ) = lim x 0 3 sin 3 x 12 sin 2 x cos 2 x . To compute this last limit we can either apply L’Hospital’s Rule again or use the fact that lim x 0 sin 3 x x = 3 , lim x 0 sin 2 x x = 2 . Consequently, lim x 0 1 cos 3 x 3 sin 2 2 x = 3 8 . 004 10.0points Determine if lim x 2 parenleftBig 2 ln( x 1) 2 x 2 parenrightBig exists, and if it does, find its value. 1. none of the other answers 2. limit = 1 correct 3. limit = 2 4. limit = 0 5. limit = + 6. limit = −∞ Explanation: After 2 ln( x 1) 2 x 2 is brought to a common denominator it can be written as f ( x ) g ( x ) = 2( x 2) 2 ln( x 1) ( x 2) ln( x 1) where f and g are functions which are differ- entiable on the interval (1 , ). In addition, lim x 2 f ( x ) = 0 , lim x 2 g ( x ) = 0 , so L’Hospital’s Rule applies. Thus lim x 2 f ( x ) g ( x ) = lim x 2 f ( x ) g ( x ) with f ( x ) = 2 2 x 1 = 2( x 2) x 1 , g ( x ) = ln( x 1) + x 2 x 1 . Since lim x 2 f ( x ) = 0 , lim x 2 g ( x ) = 0 , we still cannot evaluate the limit directly, however. But f and g are differentiable on (1 , ), so we can use L’Hospital’s Rule yet again: lim x 2 f ( x ) g ( x ) = lim x 2 f ′′ ( x ) g ′′ ( x ) with f ′′ ( x ) = 2 ( x 1) 2 , g ′′ ( x ) = 1 x 1 + 1 ( x 1) 2 . From this it follows that lim x 2 f ′′ ( x ) = 2 , lim x 2 g ′′ ( x ) = 2 . Consequently, limit = 1 . 005 10.0points
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Version 044 – K Final Exam – ben-zvi – (55600) 3 If the function f defined by f ( x ) = braceleftbigg cx + 4 , x < 2 , 4 x 2 2 , x 2 , is continuous everywhere on ( −∞ , ), what is the value of f (1)?
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