Re: 極限

看板trans_math作者 (^______^)時間18年前 (2007/06/16 10:01), 編輯推噓2(200)
留言2則, 2人參與, 最新討論串14/29 (看更多)
※ 引述《king911015 (早已放棄愛上你)》之銘言: : 1. lim (1 - 1/4)(1 - 1/9)x…x(1 - 1/X^2) = ? : x→∞ : 2. Find lim{(X^3 - 2X^2 +1)^1/3 - X} : x→∞ (x^3 - 2x^2 + 1)^(1/3) - x ((x^3-2x^2+1)^(1/3)-x)((x^3-2x^2+1)^(2/3)+((x^3 -2x^2 +1)^(1/3))(x)+ x^2) = -------------------------------------------------------------------------- (x^3 - 2x^2 + 1)^(2/3) + ((x^3 - 2x^2 + 1)^(1/3))(x) + x^2 x^3 - 2x^2 + 1 - x^3 = ------------------------------------------------------------ (x^3 - 2x^2 + 1)^(2/3) + ((x^3 - 2x^2 + 1)^(1/3))(x) + x^2 -2x^2 + 1 = ---------------------------------------------------------- (x^3 - 2x^2 + 1)^(2/3) + ((x^3 - 2x^2 + 1)^(1/3))(x) + x^2 -2 + (1/(x^2)) = ------------------------------------------------------------ (1-(2/x)+(1/(x^2)))^(2/3) +((1-(2/x)+(1/(x^2)))^(1/3))(x)+1 lim (x^3 - 2x^2 + 1)^(1/3) - x x→∞ -2 + (1/(x^2)) = lim ------------------------------------------------------------ x→∞ (1-(2/x)+(1/(x^2)))^(2/3) +((1-(2/x)+(1/(x^2)))^(1/3))(x)+1 -2 + 0 = ---------------------------------- (1-0+0)^(2/3) + (1-0+0)^(1/3) + 1 -2 -2 = ----------- = --- 1 + 1 + 1 3 : n 1 : 3. Find the limit lim Σ ------------- = ? : √(n^2 + k) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.66.173.21

06/16 10:15, , 1F
真威
06/16 10:15, 1F

06/16 12:55, , 2F
這題好像某個課本導了一公式在解
06/16 12:55, 2F
文章代碼(AID): #16SqIDlK (trans_math)
討論串 (同標題文章)
本文引述了以下文章的的內容:
2
5
完整討論串 (本文為第 14 之 29 篇):
2
9
4
21
2
11
4
25
8
23
0
1
文章代碼(AID): #16SqIDlK (trans_math)