Question

极限中x趋向于0时sinx/3xcos^3x

求极限lim(x→0)(sinx-xcosx)/(sin^3x),答案是1/3
为什么不可以这样,sinx~x,sin^3x~x^3,故原式=x-xcosx/x^3=1-cosx/x^2 而1-cosx~1/2x^2 故答案为1/2,为什么答案不对?

Answer 1

简单计算一下即可，答案如图所示

Answer 2

lim(x→0)(sinx-xcosx)/(sin^3x)
=lim(x→0)[(sinx-xcosx)]'/(sin^3x)'
=lim(x→0)(cosx-cosx+xsinx)/[3(sin^2x)cosx]
=lim(x-0) x/[3(sinxcosx)]
=lim(x-0) x/[3sin(2x)/2]
=lim(x-0) x'/[3sin(2x)/2]'
=lim(x-0) 1/[3cos(2x)]
=1/3