M2_Bk1_Sol_Ch04_E.pdf

# M2_Bk1_Sol_Ch04_E.pdf - 4 New Progress in Senior...

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© Hong Kong Educational Publishing Co. 110 New Progress in Senior Mathematics Module 2 Book 1 (Extended Part) Solution Guide 4 4 pp.133 – 140 p.133 (a) x 1.9 1.99 1.999 2 2.001 2.01 2.1 ) ( x f 0.251 58 0.250 16 0.250 02 Undefined 0.249 98 0.249 84 0.248 46 (b) 0.25 (c) 25 . 0 ) ( lim 2 = x f x p.134 (a) continuous (b) discontinuous (c) discontinuous (d) continuous pp.137 – 158 4.1 x x x x x x x 2 lim ) 6 ( lim 2 6 lim 10 10 10 = p.137 5 1 ) 10 ( 2 6 10 = = 4.2 ) 4 )( 5 ( ) 5 )( 5 ( lim 20 25 lim 5 2 2 5 + = + x x x x x x x x x p.138 9 10 4 5 5 5 4 5 lim 5 = = = x x x 4.3 (a) ) 1 )( 1 ( ) 1 )( 1 ( lim 1 1 lim 2 1 2 3 1 + + + + = + x x x x x x x x x p.139 2 3 1 1 1 ) 1 ( ) 1 ( 1 1 lim 2 2 1 = + + = + + = x x x x (b) h h h h h x x 6 6 ) 6 ( 5 lim 6 1 1 30 5 lim 6 6 + + = + + 180 ) 6 ( 30 30 lim 6 1 5 lim 6 6 = = = = h h x x 4.4 (a) 2 2 2 lim 2 + x x x p.140 4 1 2 2 2 1 2 2 1 lim ) 2 2 )( 2 ( 2 lim ) 2 2 )( 2 ( 2 ) 2 ( lim 2 2 2 2 2 2 2 lim 2 2 2 2 2 2 = + + = + + = + + = + + + = + + + + × + = x x x x x x x x x x x x x x x (b) 4 6 1 3 lim 10 + + x x x 3 4 1 10 3 4 6 10 1 3 4 6 lim 1 3 4 6 10 10 lim 1 3 4 6 4 ) 6 ( ) 1 ( ) 3 ( lim 1 3 1 3 4 6 4 6 4 6 1 3 lim 10 10 2 2 2 2 10 10 = + + = + + = + + × = + + × + = × + + + + × + + = x x x x x x x x x x x x x x x x x x x x 4.5 (a) x x x x x x x x 1 8 1 2 lim 1 8 1 2 lim + = + −∞ −∞ p.143 4 1 0 8 0 2 1 8 1 2 lim = + = + = −∞ x x x Limits and Derivatives

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111 © Hong Kong Educational Publishing Co. Limits and Derivatives (b) 2 2 2 2 7 2 4 9 lim 7 2 4 9 lim x x x x x x x x + = + 0 0 2 0 0 7 2 4 9 lim 2 2 = + = + = x x x x 4.6 x x x x x x x x x x 4 3 2 9 lim 4 3 2 9 lim 2 2 + + = + + p.144 1 0 3 0 9 4 3 2 9 lim 4 3 2 9 lim 2 2 = + + = + + = + + = x x x x x x x x x 4.7 ) 4 )( 3 )( 2 )( 1 ( ) 1 6 )( 3 2 ( lim 2 2 + + + + −∞ x x x x x x x x p.145 12 ) 0 1 )( 0 1 )( 0 1 )( 0 1 ( ) 0 6 )( 0 2 ( 4 1 3 1 2 1 1 1 1 6 3 2 lim 4 3 2 1 1 6 3 2 lim ) 4 )( 3 )( 2 )( 1 ( ) 1 6 )( 3 2 ( lim 2 2 2 2 4 2 2 = + + + + = + + + + = + + + + = + + + + = −∞ −∞ −∞ x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 4.8 ) 2 ( lim 2 x x x x + p.145 1 1 0 1 2 1 2 1 2 lim 1 2 2 lim 2 2 lim 2 2 lim 2 ) 2 ( lim 2 2 ) 2 ( lim 2 2 2 2 2 2 2 2 2 2 2 = + + = + + = + + = + + = + + = + + + = + + + + + = x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x 4.9 (a) Let . 4 n k = Then, when , n p.146 = n k 4 as well.
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