limx →0 2x^2/2sinx/2(xcosx/2+sinx/2)=limx→0 2x/(xcosx/2+sinx

蓝芸儿2022-10-04 11:39:541条回答

limx →0 2x^2/2sinx/2(xcosx/2+sinx/2)=limx→0 2x/(xcosx/2+sinx/2) 求助是怎么化过去的啊,

已提交，审核后显示！提交回复

共1条回复

文化如蕙共回答了16个问题 | 采纳率93.8%: 2x^2/[2sinx/2(xcosx/2+sinx/2)]
当x-->0时,sinx/2~x/2
将sinx/2用等价无穷小代换为x/2就行了
下面用罗比达法则OK!; 1年前

相关推荐

limx→0 2x/(xcosx/2+sinx/2)=limx→0 2/(3/2cosx/2-x/2sinx/2) 求助 limx→0 2x/(xcosx/2+sinx/2)=limx→0 2/(3/2cosx/2-x/2sinx/2) 求助是怎么化过来的 慧雪兰心1年前2: casperqing 共回答了18个问题 | 采纳率100%

lim(x->0) 2x/(xcos(x/2)+sin(x/2) ) (0/0 L's Hospital rule )
=lim(x->0) (2x)'/(xcos(x/2)+sin(x/2) )'
=lim(x->0) 2/[(cos(x/2) - (1/2)xsin(x/2) +1/2cos(x/2) )
=lim(x->0) 2/[(3/2)cos(x/2) - (1/2)xsin(x/2)]

1年前查看全部

大家在问