三角函數求极限一条 急!!

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2014-09-27 18:10:02 補充：
唔好意思, wy, 我不是有心撞題的。

我是收到 Leo 的電郵過來的。

(◕‿◕✿)

感谢wy & 知足常樂 回答。
將选知足常樂 为最佳
wy如有需要可自行刪去,不便之處請原諒

1 + tan x = 1 + x + x^3/3 + 2x^5/15 + ....... = 1 + x + x^3/3 ( neglect higher powers)
1 + sin x + 1 + x - x^3/3! + x^5/5! - ....... = 1 + x - x^3/3! ( neglect higher powers)
Based on binomial theorem,
sqrt ( 1 + x) = 1 + x/2 - x^2/8 + x^3/16 - ......
So sqrt (1 + tan x) = 1 + (x + x^3/3)/2 - (x + x^3/3)^2/8 + ......
= 1 + x + x^3/6 - (x^2 + 2x^4/3 + x^6/9)/8
= 1 + x - x^2/8 + x^3/6 + x^4/12 + ......
sqrt ( 1 + sin x) = 1 + (x - x^3/3!)/2 - (x - x^3/3!)^2/8 +......
= 1 + (x - x^3/3!)/2 - (x^2 - 2x^4/3! + x^6/36)/8 + ......
= 1 + x - x^2/8 - x^3/12 + x^4/24 + ......
so sqrt (1 + tan x) - sqrt ( 1 + sin x)
= x^3/6 + x^3/12 + x^4/12 + x^4/24 + .....
Dividing all terms by x^3 we get
1/6 + 1/12 + x/12 + x/24 + ......
So when x tends to 0, the function tends to 1/6 + 1/12 = 1/4.