K Final Exam-solutions

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

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Unformatted text preview: 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 10.0 points 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 10.0 points 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 10.0 points Find the value of lim x → 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: 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 → f ( x ) = lim x → g ( x ) = 0 . Thus L’Hospital’s Rule can be applied: lim x → f ( x ) g ( x ) = lim x → f ′ ( x ) g ′ ( x ) = lim x → 3 sin3 x 12 sin2 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 → sin3 x x = 3 , lim x → sin 2 x x = 2 . Consequently, lim x → 1 − cos 3 x 3 sin 2 2 x = 3 8 . 004 10.0 points 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 ....
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This note was uploaded on 12/05/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.

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

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