→ kkwwayne: 最後一題我只會用羅必達，謝w大 03/01 15:28
→ wayn2008: 不用謝~結果我發完文原PO刪文了XD 03/01 15:31
推 kkwwayne: 我好不容易寫完結果太慢你發文了他也刪文了 03/01 15:34
→ wayn2008: 或許推文早有強者回做法了QQ 只能說無緣XD 03/01 15:35
推 diego99: 剛打完第一題想說要推文，結果不見了XD 03/01 15:39
※ 引述《Zw111196 (無籽冰梅肉)》之銘言：
: 如題
: http://i.imgur.com/lRjVnhM.jpg
f(x)
lim ------ = 24 表示 f(x) 有 (x-1) 的因式
x>1 x-1
所以以下同理，可得
f(x) = (x-1)(x-2)(x-3)g(x)
(1)
(x-1)(x-2)(x-3)g(x)
lim ------------------- = lim [(x-2)(x-3)g(x)] = 2g(1) = 24
x>1 x-1 x>1
=> g(1) = 12
同理得 g(2) = 20 , g(3) = 30
(2) 利用牛頓差值法
g(x) = (x-1)(x-2)(x-3)Q(x) + a(x-1)(x-2) + b(x-1) + 12
g(2) = b + 12 = 20 => b = 8
g(3) = 2a + 16 +12 = 30 => a = 1
所以餘式為 x^2 + 5x + 6
(3) 利用牛頓插值法
g(x)最低次數多項式為 a(x-1)(x-2) + b(x-1) + 12 = x^2 + 5x + 6
f(x) = (x-1)(x-2)(x-3)(x^2 + 5x + 6)
: http://i.imgur.com/cbJQQwG.jpg
: 感恩~~
(1)
(x+2 - 1)^100 - 1
lim -------------------
x>-2 (x+2)
C(100,0)(-1)^100 - C(100,1)(x+2) + C(100,2)(x+2)^2 +...+ (x+2)^100 - 1
lim ------------------------------------------------------------------------
x>-2 (x+2)
lim [- C(100,1) + C(100,2)(x+2) +...+ (x+2)^99] = -100
x>-2
(2)
1 (1-x)(1 + x + x^2 + ... + x^29)
lim -----{------------------------------- - 30 }
x>1 x-1 1 - x
1 + x + x^2 + ... + x^29 - 30
= lim -------------------------------
x>1 x - 1
1-1 + (x-1) + (x^2-1) + ... + (x^29-1)
= lim ----------------------------------------
x>1 x - 1
(x-1) + (x-1)(x+1) + ... + (x-1)(x^28+x^27+...+x+1)
= lim ----------------------------------------------------
x>1 x - 1
= lim { 1 + (x+1) + (x^2+x+1) + ... + (x^28+x^27+...+x+1)}
x>1
= 1 + 2 + 3 + ... + 29
(29+1)*29
= ----------- = 435
2
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 114.42.42.129
※ 文章網址: https://www.ptt.cc/bbs/SENIORHIGH/M.1425194140.A.491.html
※ 編輯: wayn2008 (114.42.42.129), 03/01/2015 15:16:24