Help how to do this limit
2011-10-24 1:33 pm
圖片參考：http://www.mathway.com/math_image.aspx?p=SMB02LSMB03X:SMB02ZSMB03ISMB02zSMB03SMB02lSMB03SMB08%28XSMB02ESMB0310SMB02eSMB03+3XSMB02ESMB039SMB02eSMB03%29SMB02ESMB03SMB02FSMB031SMB1010SMB02fSMB03SMB02eSMB03-XSMB09?p=168?p=38
what is the limit? answer is 3/10 but how do u get it?
cannot use hospital law~~
回答 (1)
2011-10-24 4:48 pm
✔ 最佳答案
(x^10 + 3x^9)^(1/10) = [x^10(1 + 3/x)]^(1/10) = x(1 + 3/x)^(1/10).
By Binomial expansion,
x( 1 + 3/x)^(1/10) - x = x[1 + (1/10)(3/x) + (1/10)(1/10 - )/2!(3/x)^2 + (1/10) ( 1/10 - 1)(1/10 - 2)/3!(3/x)^3 + .........] - x
= x + 3/10 + (1/10)(1/10 - 1)/2!(9/x) + (1/10)(1/10 - 1)(1/10 - 2)/3! (27/x^2) + ...... - x
= 3/10 + (1/10(1/10 - 1)/2!(9/x) + (1/10)(1/10 - 1)(1/10 - 2)/3! (27/x^2) + .....
So when x tends to infinity, the expression tends to 3/10.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111024000051KK00116