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Unformatted text preview: Version 096 – K Final Exam – Hamrick – (54868) 1 This printout should have 25 questions. Multiplechoice 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 L’Hospital’s 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 L’Hospital 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 L’Hospital’s 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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 Fall '08
 schultz
 Calculus, Derivative, Differential Calculus, lim g

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