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badruddin (ssb776) – HW01 – Radin – (57410) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. HW01 is a review oF M408K material. It overlaps HW02 and HW03. 001 10.0 points Determine iF lim x →∞ b ln(7 + 2 x ) - ln(2 + 5 x ) B exists, and iF it does fnd its value. 1. limit does not exist 2. limit = ln 2 5 correct 3. limit = ln 2 7 4. limit = ln 7 2 5. limit = ln 5 2 Explanation: By properties oF logs, ln(7 + 2 x ) - ln(2 + 5 x ) = ln p 7 + 2 x 2 + 5 x P = ln p 7 /x + 2 2 /x + 5 P . But lim x →∞ 7 /x + 2 2 /x + 5 = 2 5 . Consequently, the limit exists and limit = ln 2 5 . 002 10.0 points Determine iF lim x →∞ sin 1 p 1 + 4 x 5 + 2 x 2 P exists, and iF it does, fnd its value. 1. limit = π 3 2. limit = π 4 3. limit = π 6 4. limit = π 2 5. limit = 0 correct 6. limit does not exist Explanation: Since lim x →∞ 1 + 4 x 5 + 2 x 2 = 0 , we see that lim x →∞ sin 1 p 1 + 4 x 5 + 2 x 2 P exists, and that the limit = sin 1 0 = 0 . 003 10.0 points Determine iF lim x →∞ tan 1 p 1 + 4 x 5 + 2 x 2 P exists, and iF it does, fnd its value. 1. limit = 0 correct 2. limit = π 4 3. limit = π 2 4. limit = π 3

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badruddin (ssb776) – HW01 – Radin – (57410) 2 5. limit = π 6 6. limit does not exist Explanation: Since lim x →∞ 1 + 4 x 5 + 2 x 2 = 0 , we see that lim x →∞ tan 1 p 1 + 4 x 5 + 2 x 2 P exists, and that the limit = tan 1 0 = 0 . 004 10.0 points Express the function f ( x ) = sin 2 x + 5 cos 2 x in terms of cos 2 x . 1. f ( x ) = 2 - 3 cos2 x 2. f ( x ) = - 3 + 2 cos 2 x 3. f ( x ) = 3 - 2 cos2 x 4. f ( x ) = - 2 - 3 cos 2 x 5. f ( x ) = - 2 + 3 cos 2 x 6. f ( x ) = 3 + 2 cos2 x correct Explanation: Since sin 2 x = 1 2 (1 - cos 2 x ) , cos 2 x = 1 2 (1 + cos 2 x ) , we can rewrite f ( x ) as f ( x ) = 1 2 (1 - cos 2 x ) + 5 2 (1 + cos 2 x ) . Consequently,
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