洛必达求极限(x→0)lim（x-xcosx）的导数/(x-sinx)的导数

来源：学生作业学帮网编辑：学帮网时间：2024/05/17 00:52:33

lim(x->0) (x - xcosx)/(x - sinx)
= lim(x->0) (x - xcosx)'/(x - sinx)'
= lim(x->0) [1 - (cosx - xsinx)]/(1 - cosx)
= lim(x->0) (1 - cosx + xsinx)'/(1 - cosx)'
= lim(x->0) [sinx + (sinx + xcosx)]/sinx
= lim(x->0) (2sinx + xcosx)'/(sinx)'
= lim(x->0) (2cosx + cosx - xsinx)/cosx
= lim(x->0) (3 - xtanx)
= 3 - 0
= 3

=lim(x→0)（1-cosx+xsinx）/(1-cosx)
=lim(x→0)（sinx+sinx+xcosx）/sinx
=lim(x→0)（2sinx+xcosx）/sinx
=lim(x→0)（2cosx+cosx-xsinx）/cosx
=lim(x→0)（3cosx-xsinx）/cosx
=3

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