第6题第7个？
1个回答
展开全部
(7)
x->0
ln(1+tanx)
=ln[ 1+ x+ (1/3)x^3 +o(x^3) ]
=[x+ (1/3)x^3 +o(x^3)]-(1/2)[x+ (1/3)x^3 +o(x^3)]^2
+(1/3)[x+ (1/3)x^3 +o(x^3)]^3 +o(x^3)
=[x+ (1/3)x^3 +o(x^3) ] - (1/2)[x^2 +o(x^3)] +(1/3)[x^3 +o(x^3) ]
=x -(1/2)x^2 + (2/3)x^3 +o(x^3)
ln(1+sinx)
=ln[1+ x- (1/6)x^3 +o(x^3)]
=[x- (1/6)x^3 +o(x^3)]-(1/2)[ x- (1/6)x^3 +o(x^3)]^2
+(1/3)[ x- (1/6)x^3 +o(x^3)]^3 +o(x^3)
=[x- (1/6)x^3 +o(x^3)] - (1/2)[x^2 +o(x^3)] +(1/3)[x^3 +o(x^3) ]
=x -(1/2)x^2 + (1/6)x^3 +o(x^3)
ln[(1+tanx)/(1+sinx) ]
=ln(1+tanx) -ln(1+sinx)
=(2/3 -1/6 )x^3 +o(x^3)
= (1/2)x^3 +o(x^3)
//
lim(x->0) [(1+tanx)/(1+sinx) ]^(1/x^3)
=lim(x->0) e^{ ln[(1+tanx)/(1+sinx) ] /x^3 }
=lim(x->0) e^[ (1/2)x^3 /x^3 ]
=e^(1/2)
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
