Assignment3Solution(1804)

Assignment3Solution(1804) - MATH1804/Solution to...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1804: University Mathematics A Solution to assignment 3 08/10/2008 1. Solution : (a) lim x ! 1 5 x 4 4 x 2 1 10 x 9 x 3 = lim x ! 1 20 x 3 8 x 1 27 x 2 = 3 7 ; (b) lim x ! 1 2 x 2 (3 x + 1) p x + 2 x 1 = lim x ! 1 4 x 9 2 x 1 = 2 1 2 x 1 = 2 1 = 1; (c) lim x !1 x tan 1 x = lim x !1 tan (1 =x ) 1 =x = lim x !1 cos 2 (1 =x ) ± x 2 ± x 2 = lim x !1 cos 2 (1 =x ) = 1; (d) lim x ! 0 sin 9 x tan 27 x = lim x ! 0 9 cos 9 x 27 cos 2 27 x = lim x ! 0 1 3 cos 9 x cos 2 27 x = 1 3 ; (e) lim x ! 0 sin x cos x x x 2 = lim x ! 0 cos 2 x sin 2 x 1 2 x = 1 . 2. Solution: Let f ( z ) = sin z for any z 2 R . For any x , y 2 R , since f is continuous and di±erentiable on the whole real line, by Mean Value Theorem, there exists a number c lying between x and y , such that f ( x ) f ( y ) = f 0 ( c ) ( x y ) , or equivalently, sin x sin y = (cos c ) ( x y ) . Therefore we have j sin x sin y j = j cos c jj x y j . Since 1 ² cos c ² 1 , j cos c j ² 1 , which implies that j sin x sin y j ² j x y j . 1
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This note was uploaded on 10/14/2010 for the course MATH math1804 taught by Professor Prof during the Spring '08 term at HKU.

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Assignment3Solution(1804) - MATH1804/Solution to...

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