SampleProblems (2) - AUBG Spring 2009, MAT 103 “Calculus...

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Unformatted text preview: AUBG Spring 2009, MAT 103 “Calculus I” Final Exam Sample Problems Problem 1. Evaluate the limit. (a) lim x →− 1 3 √ x + 1 x + 1 (b) lim x → sin 3 x sin 4 x (c) lim x →∞ x 2 + 1 √ x 4 + 1 + x 2 (d) lim x →∞ x ( radicalbig x 2 + 1- x ) Hint: You may try to rationalize the numerator in part (d). Answers. (a) 1 / 3 (b) 3 / 4 (c) 1 / 2 (d) 1 / 2 Solutions. (a) lim x →− 1 3 √ x + 1 x + 1 = lim x →− 1 ( 3 √ x + 1)( 3 √ x 2- 3 √ x + 1) ( x + 1)( 3 √ x 2- 3 √ x + 1) = lim x →− 1 x + 1 ( x + 1)( 3 √ x 2- 3 √ x + 1) = lim x →− 1 1 3 √ x 2- 3 √ x + 1 = 1 3 p (- 1) 2- 3 √- 1 + 1 = 1 3 (b) lim x → sin 3 x sin 4 x = lim x → (sin 3 x ) /x (sin 4 x ) /x = lim x → (sin 3 x ) /x lim x → (sin 4 x ) /x = 3 4 (c) lim x →∞ x 2 + 1 √ x 4 + 1 + x 2 = lim x →∞ x 2 √ x 4 + x 2 = lim x →∞ x 2 x 2 + x 2 = 1 2 (d) lim x →∞ x ( p x 2 + 1- x ) = lim x →∞ x ( √ x 2 + 1- x )( √ x 2 + 1 + x ) √ x 2 + 1 + x = lim x →∞ x √ x 2 + 1 + x = lim x →∞ x √ x 2 + x = lim x →∞ x x + x = 1 2 Problem 2. Express the limit as a derivative and evaluate. (a) lim h → 1 10 √ 1 + h- 1 h (b) lim h → √ cos h- 1 h (c) lim x → π/ 4 √ tan x- 1 x- π/ 4 Answers. (a)- 1 / 10 (The derivative of f ( x ) = x − 1 / 10 at x = 1.) (b) 0 (The derivative of f ( x ) = √ cos x at x = 0.) (c) 1 (The derivative of f ( x ) = √ tan x at x = π/ 4.) Solutions. (a) lim h → 1 10 √ 1 + h- 1 h = lim h → f ( a + h )- f ( a ) h = f ′ ( a ), where f ( x ) = 1 10 √ x = x − 1 / 10 and a = 1. Since f ′ ( x ) =- 1 10 x − 11 / 10 , the limit is equal to- 1 10 1 − 11 / 10 =- 1 10 . (b) lim h → √ cos h- 1 h = lim h → f ( a + h )- f ( a ) h = f ′ ( a ), where f ( x ) = √ cos x and a = 0. Since f ′ ( x ) =- sin x 2 √ cos x , the limit is equal to- sin 0 2 √ cos 0 = 0. (c) lim x → π/ 4 √ tan x- 1 x- π/ 4 = lim x → π/ 4 f ( x )- f ( a ) x- a = f ′ ( a ), where f ( x ) = √ tan x and a = π/ 4. Since f ′ ( x ) = 1 2 √ tan x · 1 cos 2 x , the limit is equal to 1 2 p tan π/ 4 · 1 cos 2 π/ 4 = 1. Problem 3. Find y ′ . (a) y = x 3 cos 4 x (b) y = √ x 3 + x x 2 + x + 1 (c) y = (1 + tan 2 x ) 5 (d) y = 3 radicalBig 1 + √ 1 + x Solutions. (a) y ′ = 3 x 2 · cos 4 x + x 3 · 4 cos 3 x · (- sin x ) = 3 x 2 cos 4 x- 4 x 3 cos 3 x sin x (b) y ′ = 1 2 √ x 3 + x · (3 x 2 + 1) · ( x 2 + x + 1)- √ x 3 + x · (2 x + 1) ( x 2 + x + 1) 2 =- x 4 + x 3- x + 1 2 √ x 3 + x ( x 2 + x + 1) 2 (c) y ′ = 5(1 + tan 2 x ) 4 · 2 tan x · 1 cos 2 x = 10 tan x (1 + tan 2 x ) 4 cos 2 x (d) y ′ = 1 3 (1 + √ 1 + x ) − 2 / 3 · 1 2 √ 1 + x = 1 6 √ 1 + x ( 3 p 1 + √ 1 + x ) 2 Problem 4. Use implicit differentiation to find an equation for the tangent line to the curve at the given point....
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This note was uploaded on 12/14/2009 for the course BUSINESS bus 210 taught by Professor Levine during the Spring '09 term at American University in Bulgaria.

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SampleProblems (2) - AUBG Spring 2009, MAT 103 “Calculus...

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