最后一次串讲（修改于1月4号）

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I n → +∞, x → x0, x → +∞, 1 n = 1 103 limx→0+ xx 1 = 1 ⇒ limn→+∞ n 2 /18
II sin x limx→0 x limx→∞ = 1 1 x 1 + = e 4224 lim sinx ∼ x, tanx ∼ x, x tanx − sinx limx → 0 x3 tan x−sinx x3 ,: = lim x→0 x − x = 0. x3 3 /18
Page 1 and 2: Tiao Lu Email: tlu@math.pku.edu.cn
Page 3 and 4: I n → +∞, x → x0, x → +
Page 5: I n → +∞, x → x0, x → +
Page 9 and 10: II sin x limx→0 x limx→∞ =
Page 11 and 12: II sin x limx→0 x limx→∞ =
Page 13 and 14: Problem limx→1x 1 x−1 t = x −
Page 15 and 16: Problem limx→0 1 1 x − ex −1
Page 21 and 22: 1 f (x) = 2 1 x = 1 x−1 +1 (A)(B)
Page 23 and 24: f ′ (−x)f ′ (x)−x. f (−x)
Page 25 and 26: f ′ (−x)f ′ (x)−x. f (−x)
Page 27 and 28: f ′ (−x)f ′ (x)−x. f (−x)
Page 29 and 30: f ′ (−x)f ′ (x)−x. f (−x)
Page 31 and 32: y = f (x)f (x) = f (0) + x + α(x),
Page 33 and 34: ” ∞ ∞ ”, ”0 0 ”, ”∞
Page 35 and 36: ” ∞ ∞ ”, ”0 0 ”, ”∞
Page 37 and 38: ” ∞ ∞ ”, ”0 0 ”, ”∞
Page 39 and 40: ” ∞ ∞ ”, ”0 0 ”, ”∞
Page 41 and 42: Problem : ln(1 + x) < x √ 1+
Page 43 and 44: Problem : ln(1 + x) < x √ 1+
Page 45 and 46: Problem : sinx > 2 π πxx ∈
Page 47 and 48: e.g. b a f (x)dx = b a f (t)dt u-
Page 49 and 50: Problem a 0 √ a 2 − x 2
Page 51 and 52: Problem 1 1+ 3√ 1+x dx u =
Page 53 and 54: Problem a, bf (x) = x3 + ax2 +
Page 55 and 56: ln(x + 1) < f (x) = ln(x + 1) − x

最后一次串讲（修改于1月4号）

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