※ 引述《JackieYu (聽不到)》之銘言： : 就快轉系考了 : 心急如焚請好心人幫忙 : 拜託高手賜教 Orz(泣) : http://homepage.ntu.edu.tw/~b92501063/a.jpg 1. (e^x - 1)^3 lim ------------------- x->0 (x -2)e^x + x + 2 [3(e^x - 1)^2]*(e^x) = lim ---------------------- x->0 e^x + (x - 2)e^x + 1 [6(e^x - 1)e^x]*(e^x) + [3(e^x - 1)^2]*(e^x) = lim --------------------------------------------- x->0 e^x + e^x + (x - 2)e^x [6(e^x - 1)e^x] + [3(e^x - 1)^2] = lim ---------------------------------- x->0 x [6(e^x)(e^x) + 6(e^x - 1)e^x] + [6(e^x - 1)e^x] = lim ------------------------------------------------- = 6 x->0 1 x f(t) 4. 設連續函數f(x)滿足 6 + ∫ ------ dt = 2*((x)^(1/2)) a t^2 則 f(x) = _________ , a = ________ x f(t) 解:令 F(x) = 6 + ∫ ------ dt = 2*((x)^(1/2)) a t^2 a f(t) 則 F(a) = 6 + ∫ ------ dt = 2*((a)^(1/2)) a t^2 => 6 = 2*((a)^(1/2)) => 3 = (a)^(1/2) => a = 9 f(x) F'(x) = ------ = 2*(1/2)*((x)^(-1/2)) = (x)^(-1/2) x^2 所以 f(x) = (x)^(3/2) 3 ln(x+1) 6. ∫ --------- dx 1 x^2 ln(x+1) |3 3 1 1 = - --------- | - ∫ (- ---)*(-----) dx  x |1 1 x x+1 ln4 3 1 1 = - ----- + ln2 + ∫ (--- - -----) dx 3 1 x x+1 1 | x | |3 = ---ln2 + ln|-----| | 3 | x+1 | |1 1 3 1 = ---ln2 + ln--- - ln--- 3 4 2 4 3 2 = ---ln2 + ln--- = - ---ln2 + ln3 3 4 3 --

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