HW1-510-soln

# 2 h 4 16 h lim h h 3 8h 2 f 1 x 2 x 1 1 2 x 1 2 2 x 1

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0 2 h 4 16 h = lim h 0 h 3 8h 2 24h 32 = 32 (f) (1) x – 2 x 1 1 2 = x 1 2 2 x 1  x 1 2 for any  0 , there exists such that = min { , 3 4 } , if x – 1  , we have x – 1 and

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x 1  1 4 ,so x – 2 x 1 1 2 = x 1 2 2 x 1  x 1 2 x 1 2 2 1 4 1  1 4 1 2 = 8 45 x 1 2 8 45  , that is, lim x 1 x x 1 = 1 2 (2) lim x 1 x x 1 = lim x 1 x lim x 1 x 1 = 1 2 3.52 (a) (b) lim x 2 0 f x = lim x 2 0 2x 5 = 9 , lim x 2 0 f x = lim x 2 0 2x 5 = 9 , so lim x 2 f x = 9 (c) lim x − 3 0 f x = lim x − 3 0 3x 1 =− 10 , lim x − 3 0 f x = lim x − 3 0 3x 1 =− 10 , so lim x − 3 f x = 9 (d) lim x 0 0 f x = lim x 0 0 2x 5 = 5 (e) lim x 0 0 f x = lim x 0 0 3x 1 =− 1 (f) lim x 0 0 f x ≠ lim x 0 0 f x  ⇒ lim x 0 f x Not exist. 3.71 (a) lim x 0 e ax e bx x = lim x 0 e ax 1 x lim x 0 e bx 1 x = lim x 0 e ax 1 − a x − a − lim x 0 e bx 1 − b x − b = b a (b) lim x 0 a x b x x = lim x 0 e x ln a e x ln b x , from (a), we get lim x 0 e x ln a e x ln b x = ln a ln b = ln a b
(c) lim x 0 tanh ax x = lim x 0 e ax e ax e ax e ax x = lim x 0 e ax e ax x 1 e ax e ax = lim x 0 e ax e ax x lim x 0 1 e ax e ax =[ a −− a ]⋅ 1 1 1 = a 3.75 (a) discontinuous. lim x 0 f x = lim x 0 sin x x = 1 0 = f 0 (b)continuous. lim x 0 0 f x = lim x 0 0 x x = 0 , lim x 0 0 f x = lim x 0 0 x x = 0 so lim x 0 0 f x = lim x 0 0 f x = 0 = f 0 (c) continuous.
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