Calculus 2-A3 - Calculus II Calculus: Early Transcendental...

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Calculus II Calculus: Early Transcendental Functions , 4th ed., by Ron Larson, Robert Hostetler, and Bruce H. Edwards (Boston: Houghton Mifflin, 2007; ISBN-10: 0-618-60624-6). Written Assignment 3 Complete the following textbook exercises for sections 8.7 and 8.8, and submit them to your mentor for correction and grading. Show all calculations. SECTION 8.7 Exercises Do exercises 6, 8, 10, 14, 22, 32, 38, 42, 44, 52, and 54 on page 574 of the textbook. #6 In Exercises 5– 10, evaluate the limit ( a) using techniques from Chapters 2 and 4 and ( b) using L’Hôpital’s Rule. (a) lim (2x-3)(x+1) x->-1 x + 1 = lim (2x-3) x->-1 = -5 (b) L’Hôpital’s Rule lim (d/dx)(2x 2 - x -3)(x+1) x->-1 (d/dx) (x + 1) = lim (4x - 1) x->-1 1 = -5 #8 (a) Using trigonometric limits. .. =2[lim x =>0 [sin(x)/(x)] =2(1) =2 (b) L’Hôpital’s Rule = lim x =>0 [sin(4*0)/(2*0)] = lim x =>0 [sin(0)/(0)] = lim x =>0 [(0)/(0)] = lim x =>0 [4cos(4x)/(2)] = lim x =>0 [2cos(4x)]
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= lim x =>0 [2cos(4*0)] = 2cos(0) =2 #10 (a) Using trigonometric limits. .. = lim x =>∞ [2x/(4x 2 +x)] + lim x =>∞ [1/(4x 2 +x)] = lim x =>∞ [2/(4x 2 +1)] + lim x =>∞ [1/(4x 2 +x)] = 0/4 = 0 (b) L’Hôpital’s Rule = lim x =>∞ (2x + 1) lim x =>∞ (4x 2 +x) = ∞ indeterminate = lim x =>∞ (2x + 1) (4x 2 +x) = lim x =>∞
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Calculus 2-A3 - Calculus II Calculus: Early Transcendental...

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