Int calc hw 1

# Int calc hw 1 - pokharel(yp624 HW01 Radin(57410 This...

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pokharel (yp624) – 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(3 + 4 x ) - ln(8 + 5 x ) B exists, and iF it does fnd its value. 1. limit = ln 5 4 2. limit = ln 3 8 3. limit = ln 8 3 4. limit = ln 4 5 correct 5. limit does not exist Explanation: By properties oF logs, ln(3 + 4 x ) - ln(8 + 5 x ) = ln p 3 + 4 x 8 + 5 x P = ln p 3 /x + 4 8 /x + 5 P . But lim x →∞ 3 /x + 4 8 /x + 5 = 4 5 . Consequently, the limit exists and limit = ln 4 5 . 002 10.0 points Determine iF lim x →∞ sin 1 p 1 + x 3 + 2 x P exists, and iF it does, fnd its value. 1. limit = 0 2. limit = π 6 correct 3. limit = π 4 4. limit does not exist 5. limit = π 3 6. limit = π 2 Explanation: Since lim x →∞ 1 + x 3 + 2 x = 1 2 , we see that lim x →∞ sin 1 p 1 + x 3 + 2 x P exists, and that the limit = sin 1 1 2 = πover 6 . 003 10.0 points Determine iF lim x →∞ tan 1 p 3 + 2 x 2 1 + 4 x P exists, and iF it does, fnd its value. 1. limit = π 4 2. limit = π 2 correct 3. limit = π 3 4. limit = π 6

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pokharel (yp624) – HW01 – Radin – (57410) 2 5. limit does not exist 6. limit = 0 Explanation: Since 3 + 2 x 2 1 + 4 x -→ ∞ as x → ∞ , we see that lim x →∞ tan 1 p 3 + 2 x 2 1 + 4 x P exists, and that the limit = tan 1 = π 2 . 004 10.0 points Express the function f ( x ) = 5 sin 2 x - cos 2 x in terms of cos 2 x . 1. f ( x ) = 3 + 2 cos2 x 2. f ( x ) = - 3 - 2 cos 2 x 3. f ( x ) = 2 + 3 cos2 x 4. f ( x ) = - 2 - 3 cos 2 x 5. f ( x ) = 2 - 3 cos2 x correct 6. f ( x ) = 3 - 2 cos2 x 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 ) = 5 2 (1 - cos 2 x ) - 1 2 (1 + cos 2 x ) . Consequently,
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Int calc hw 1 - pokharel(yp624 HW01 Radin(57410 This...

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