lim (tanx-x)/(x-sinx)(x->0)利用洛必达法则
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/11 05:49:21
![lim (tanx-x)/(x-sinx)(x->0)利用洛必达法则](/uploads/image/z/615047-23-7.jpg?t=lim+%28tanx-x%29%2F%28x-sinx%29%28x-%3E0%29%E5%88%A9%E7%94%A8%E6%B4%9B%E5%BF%85%E8%BE%BE%E6%B3%95%E5%88%99)
xN@_`hNitoA΅VVƝ# [Q@\k/#`6q9N䶃NnRI4PN**&ZzP
^wh6YN&z`K rZz5P~*dFFhs_;[fvh]Noc8M?r"
M4łzя ^PU=~p/'&%|lAT֡`u4 Pb Jԁ\x%-Fҙ
lim (tanx-x)/(x-sinx)(x->0)利用洛必达法则
lim (tanx-x)/(x-sinx)(x->0)利用洛必达法则
lim (tanx-x)/(x-sinx)(x->0)利用洛必达法则
x->0时,
lim (tanx-x)/(x-sinx)
=lim[(1/cos²x)-1]/(1-cosx)
=lim(1-cos²x)/[cos²x(1-cosx)]
=lim(1+cosx)/cos²x
=2
过程见下图: