=∫（x^3到x^2）(sin(xt) d(xt) /(xt) ) / x^2
令積分中的(xt) =u,則
=∫（x^3到x^2）(sinu du /u ) / x^2
利用洛比達(dá)法則,
= lim [2x·sin(x^2) /(x^2) - 3x^2·sin(x^3) /(x^3) ] /(2x)
= lim [2·sin(x^2) - 3sin(x^3)] /(2x^2)
= lim sin(x^2)/(x^2) - lim 3sin(x^3)/(2x^2) //因為減號兩邊各部分的極限都不是無窮大,故可以分開來求
= lim sin(x^2)/(x^2) - lim 3(x^3)/(2x^2) //等價無窮小代換
=1-0
=1lim [2x·sin(x^2)/(x^2) - 3x^2·sin(x^3)/(x^3) ] /(2x)到這步= lim [2·sin(x^2) - 3sin(x^3)] /(2x^2)好像3sin(x^3)少個x項把，還是我弄錯了，你在看看被哦我明白了，這樣對謝謝啊