K Final Exam-solutions

K Final Exam-solutions - Version 096 K Final Exam Hamrick...

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Unformatted text preview: Version 096 K Final Exam Hamrick (54868) 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 8 2 4 Use the graph to determine lim x 3 f ( x ) . 1. limit = 12 2. limit = 7 3. limit does not exist correct 4. limit = 4 5. limit = 3 Explanation: From the graph it is clear the f has a left hand limit at x = 3 which is equal to 3; and a right hand limit which is equal to 2. Since the two numbers do not coincide, the limit does not exist . 002 10.0 points Find the value of lim x 3 2 x 6 x 3 if the limit exists. 1. limit = 6 3 2. limit = 12 3. limit = 3 3 4. limit does not exist 5. limit = 2 3 6. limit = 4 3 correct Explanation: Since x 3 = ( x + 3)( x 3) , we can rewrite the given expression as 2( x + 3)( x 3) x 3 = 2( x + 3) for x negationslash = 3. Thus lim x 3 2 x 6 x 3 = 4 3 . 003 10.0 points Determine lim x x 4 3 x 3 + 6 . 1. limit = 7 2. limit = 0 3. limit = correct 4. limit = 5. limit = 3 6. none of the other answers Explanation: 004 10.0 points Version 096 K Final Exam Hamrick (54868) 2 Determine if lim x (ln x ) 2 6 x + 4 ln x exists, and if it does, find its value. 1. limit = 2. limit = 0 correct 3. limit = 10 4. none of the other answers 5. limit = 6 6. limit = Explanation: Set f ( x ) = (ln x ) 2 , g ( x ) = 6 x + 4 ln x . Then f, g have derivatives of all orders and lim x f ( x ) = , lim x g ( x ) = . Thus LHospitals Rule applies: lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) . But f ( x ) = 2 ln x x , g ( x ) = 6 + 4 x , so lim x f ( x ) g ( x ) = lim x 2 ln x 6 x + 4 . We need to apply LHospital once again, for then lim x 2 ln x 6 x + 4 = lim x 2 x 6 = 0 . Consequently, the limit exists and lim x (ln x ) 2 6 x + 4 ln x = 0 . 005 10.0 points Determine lim x 4 x tan 1 (5 x ) . 1. limit = 4 5 correct 2. limit = 0 3. limit = 4 4. limit = 5 4 5. limit = 1 5 6. limit does not exist Explanation: Since the limit has the form lim x 4 x tan 1 (5 x ) = , we use LHospitals Rule with f ( x ) = 4 x, g ( x ) = tan 1 (5 x ) . For then lim x f ( x ) g ( x ) = lim x f ( x ) g ( x ) = lim x 4(1 + (5 x ) 2 ) 5 . Consequently, limit = 4 5 . 006 10.0 points Determine which (if any) of the following functions is not continuous at x = 5. 1. all continuous at x = 5 Version 096 K Final Exam Hamrick (54868) 3 2. f ( x ) = braceleftbigg | x 5 | x negationslash = 5 x = 5 3. f ( x ) = braceleftBigg 10 2 x 5 x negationslash = 5 2 x = 5 4. f ( x ) = braceleftBigg 1 x 5 x negationslash = 5 5 x = 5 correct 5. f ( x ) =...
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This note was uploaded on 04/30/2011 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.

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K Final Exam-solutions - Version 096 K Final Exam Hamrick...

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