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M408D Quest Homework 1-solutions

M408D Quest Homework 1-solutions - hernandez(ah29758 M408D...

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hernandez (ah29758) – M408D Quest Homework 1 – pascaleff – (54550) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Determine if the limit lim x 1 x 8 1 x 7 1 exists, and if it does, find its value. 1. limit = −∞ 2. limit = 3. limit = 8 7 correct 4. none of the other answers 5. limit = 7 8 Explanation: Set f ( x ) = x 8 1 , g ( x ) = x 7 1 . Then lim x 1 f ( x ) = 0 , lim x 1 g ( x ) = 0 , so L’Hospital’s rule applies. Thus lim x 1 f ( x ) g ( x ) = lim x 1 f ( x ) g ( x ) . But f ( x ) = 8 x 7 , g ( x ) = 7 x 6 . Consequently, limit = 8 7 . 002 10.0points Determine if the limit lim x → −∞ (1 5 x ) 1 6 x exists, and if it does, find its value. 1. none of the other answers 2. limit = 1 correct 3. limit = −∞ 4. limit = 0 5. limit = e 1 5 6. limit = e 5 7. limit = Explanation: Since the limit has the form lim x → −∞ (1 5 x ) 1 6 x = 0 , we first take logs and evaluate lim x → −∞ ln braceleftBig (1 5 x ) 1 6 x bracerightBig . But ln braceleftBig (1 5 x ) 1 6 x bracerightBig = ln(1 5 x ) 6 x . By L’Hospital’s Rule, therefore, lim x → −∞ ln braceleftBig (1 5 x ) 1 6 x bracerightBig = lim x → −∞ 5 6 parenleftBig 1 1 5 x parenrightBig = 0 . Consequently, the limit exists and limit = e 0 = 1 . 003 10.0points Determine if the improper integral I = integraldisplay 6 2 ( x + 4) 2 dx converges, and if it does, compute its value.
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hernandez (ah29758) – M408D Quest Homework 1 – pascaleff – (54550) 2 1. I = 3 7 2. I = 1 5 correct 3. I = 1 5 4. I doesn’t converge 5. I = 2 11 Explanation: The integral is improper because of the infi- nite interval of integration. To overcome this we set I = lim t → ∞ integraldisplay t 6 2 ( x + 4) 2 dx whenever the limit on the right hand side exists. Now integraldisplay t 6 2 ( x + 4) 2 dx = bracketleftBig 2 x + 4 bracketrightBig t 6 = 2 t + 4 + 1 5 . Since lim t → ∞ 2 t + 4 = 0 , it follows that the improper integral converges and that I = integraldisplay 6 2 ( x + 4) 2 dx = 1 5 .
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