HW01-solutions - wei(jw35975 HW01 kalahurka(55230 This...

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wei (jw35975) – HW01 – kalahurka – (55230) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Determine lim x 0 braceleftBig 1 x 2 + x 1 x bracerightBig . 1. limit = 1 2. limit = 1 2 3. limit = 1 2 4. limit = 1 correct 5. limit = 1 3 6. limit = 1 3 Explanation: After simplification we see that 1 x 2 + x 1 x = 1 ( x + 1) x ( x + 1) = 1 x + 1 , for all x negationslash = 0. Thus limit = lim x 0 1 x + 1 = 1 . 002 10.0points Determine if lim h 0 f ( x + h ) f ( x ) h exists when f ( x ) = 2 x 2 + 4 x + 4, and if it does, find its value. 1. limit = 2 x + 4 2. limit does not exist 3. limit = 5 x + 4 4. limit = 4 x + 4 correct 5. limit = 3 x + 4 6. limit = 6 x + 4 Explanation: Since f ( x + h ) f ( x ) = 2( x + h ) 2 + 4( x + h ) + 4 = 2 x 2 + (4 x + 4) h + 2 h 2 + 4 , we see that f ( x + h ) f ( x ) h = 2 h + 4 x + 4 . On the other hand, lim h 0 (2 h + 4 x + 4) = 4 x + 4 . Consequently, lim h 0 f ( x + h ) f ( x ) h exists when f ( x ) = 2 x 2 + 4 x + 4, and has limit = 4 x + 4 . 003 10.0points Find the value of lim x 5 2 x 10 x 5 if the limit exists. 1. limit = 6 5 2. limit = 3 5 3. limit = 4 5 correct 4. limit does not exist
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wei (jw35975) – HW01 – kalahurka – (55230) 2 5. limit = 2 5 6. limit = 20 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 . 004 10.0points Determine lim x 0 x 1 x 2 ( x + 4) . 1. limit = 1 4 2. none of the other answers 3. limit = 4. limit = −∞ correct 5. limit = 0 6. limit = 1 Explanation: Now lim x 0 x 1 = 1 . On the other hand, x 2 ( x + 4) > 0 for all small x , both positive and negative, while lim x 0 x 2 ( x + 4) = 0 . Consequently, limit = −∞ . keywords: evaluate limit, rational function 005 10.0points Determine lim x → ∞ 6 x 2 2 x + 8 2 + 7 x 3 x 2 . 1. limit = 2 correct 2. limit = 3. limit = 0 4. none of the other answers 5. limit = 1 Explanation: Dividing the numerator and denominator by x 2 we see that 6 x 2 2 x + 8 2 + 7 x 3 x 2 = 6 2 x + 8 x 2 2 x 2 + 7 x 3 . On the other hand, lim x → ∞ 1 x = lim x → ∞ 1 x 2 = 0 . By Properties of limits, therefore, the limit = 2 . 006 10.0points Determine if lim x → ∞ x parenleftBig radicalbig 9 x 2 + 7 3 x parenrightBig exists, and if it does, find its value. 1. limit = 1 2. limit = 5 6
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wei (jw35975) – HW01 – kalahurka – (55230) 3 3. limit does not exist 4. limit = 7 6 correct 5. limit = 2 3 6. limit = 4 3 Explanation: After rationalization we see that x parenleftBig radicalbig 9 x 2 + 7 3 x parenrightBig = x parenleftbigg 9 x 2 + 7 9 x 2 9 x 2 + 7 + 3 x parenrightbigg = 7 x 9 x 2 + 7 + 3 x = 7 radicalbigg 9 + 7 x 2 + 3 . On the other hand, lim x → ∞ radicalbigg 9 + 7 x 2 = 3 . Consequently, lim x → ∞ x parenleftBig radicalbig 9 x 2 + 7 3 x parenrightBig exists and has limit = 7 6 .
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