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Question 41
Thank you for your help.
Easy limit
2013-11-17 4:33 am
回答 (2)
2013-11-17 5:04 am
2013-11-17 5:43 pm
e^(x^3) = 1 + x^3 + x^6/2! + x^9/3! + ........
so e^(x^2) - 1 = x^3 + x^6/2! + x^9/3! + ....... = x^3(1 + x^3/2! + x^6/3! + ......)
sin x = x - x^3/3! + x^5/5! - x^7/7! + ......
so x - sin x = x^3/3! - x^5/5! + x^7/7! - ...... = x^3(1/3! - x^2/5! + x^4/7! - .....)
so [ e^(x^3) - 1 ]/(x - sin x)
= (1 + x^3/2! + x^6/3! + ....)/(1/3! - x^2/5! + x^4/7! - .....)
so limit when x tends to 0 = 1/(1/3!) = 3! = 6.
so e^(x^2) - 1 = x^3 + x^6/2! + x^9/3! + ....... = x^3(1 + x^3/2! + x^6/3! + ......)
sin x = x - x^3/3! + x^5/5! - x^7/7! + ......
so x - sin x = x^3/3! - x^5/5! + x^7/7! - ...... = x^3(1/3! - x^2/5! + x^4/7! - .....)
so [ e^(x^3) - 1 ]/(x - sin x)
= (1 + x^3/2! + x^6/3! + ....)/(1/3! - x^2/5! + x^4/7! - .....)
so limit when x tends to 0 = 1/(1/3!) = 3! = 6.
收錄日期: 2021-04-25 22:43:27
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