Midterm Exam 2 Solutions

Midterm Exam 2 Solutions - MATH150 Midterm Exam 2 Solutions...

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MATH150 Midterm Exam 2 Solutions 11/18/05 1. (12 marks) Determine the following limits: (a) lim x →∞ ln(ln x ) ln 2 x (b) lim x 0 sinh( x ) - x x 3 (c) lim x π/ 2 - (cos x ) cos x Solution : (a) lim x →∞ ln(ln x ) ln 2 x = lim x →∞ 1 ln x · 1 x 2 ln x · 1 x = lim x →∞ 1 2 ln 2 x = 0. (b) lim x 0 sinh x - x x 3 0 0 = lim x 0 cosh x - 1 3 x 2 0 0 = lim x 0 sinh x 6 x 0 0 = lim x 0 cosh x 6 = 1 6 . (c) Let’s first compute: lim x π/ 2 - cos x ln(cos x ) = lim x π/ 2 - ln(cos x ) sec x = lim x π/ 2 - 1 cos x · ( - sin x ) sec x tan x = lim x π/ 2 - - tan x sec x tan x = lim x π/ 2 - - 1 sec x = lim x π/ 2 - - cos x = 0 Thus lim x π/ 2 - (cos x ) cos x = e lim x π/ 2 - cos x ln(cos x ) = e 0 = 1. 2. (12 marks) What is the maximal area a right triangle can have whose hypotenuse has a length of 2 cm? Carefully justify your answer! Solution : Let the lengths of the sides of the triangle be x , y and 2. The area of the triangle is A = 1 2 xy , where x 2 + y 2 = 4 y = 4 - x 2 . Thus we have to maximize A ( x ) = 1 2 x 4 - x 2 , defined on the interval [0 , 2]. A
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Midterm Exam 2 Solutions - MATH150 Midterm Exam 2 Solutions...

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