∫（x^2到x）(sin(xt)dt/t)/x^2,求极限x趋于0,我算等于-1/2,可是答案是1,

Question

∫（x^2到x）(sin(xt) dt/t)/x^2,求极限x趋于0,
我算等于-1/2,可是答案是1,

Answer 1

=∫（x^3到x^2）(sin(xt) d(xt) /(xt) ) / x^2
令积分中的(xt) =u,则
=∫（x^3到x^2）(sinu du /u ) / x^2
利用洛比达法则,
= 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) //因为减号两边各部分的极限都不是无穷大,故可以分开来求
= lim sin(x^2)/(x^2) - lim 3(x^3)/(2x^2) //等价无穷小代换
=1-0
=1