求limx趋近于0ln(1 5x)arcsin3x/cos2x-1
1个回答
展开全部
lim(x->晌御0) ln(1+5x)arcsin(3x)/[cos(2x)-1]
=lim(x->0) [ln(1+5x)arcsin(3x)]'/[cos(2x)-1]'
=lim(x->0) [5arcsin(3x)/(1+5x)+3ln(1+5x)/√(1-9x^2)]/[-2sin(2x)]
=lim(x->0) [5arcsin(3x)/(1+5x)+3ln(1+5x)/√(1-9x^2)]'/[-2sin(2x)]'
=lim(x->0) {[25arcsin(3x)/(1+5x)^2-30/吵渣[(1+5x)√升谨悄(1-9x^2)]-27xln(1+5x)/[(1-9x^2)√(1-9x^2)]}/[8cos(2x)]
=-30/8
=-15/4
=lim(x->0) [ln(1+5x)arcsin(3x)]'/[cos(2x)-1]'
=lim(x->0) [5arcsin(3x)/(1+5x)+3ln(1+5x)/√(1-9x^2)]/[-2sin(2x)]
=lim(x->0) [5arcsin(3x)/(1+5x)+3ln(1+5x)/√(1-9x^2)]'/[-2sin(2x)]'
=lim(x->0) {[25arcsin(3x)/(1+5x)^2-30/吵渣[(1+5x)√升谨悄(1-9x^2)]-27xln(1+5x)/[(1-9x^2)√(1-9x^2)]}/[8cos(2x)]
=-30/8
=-15/4
本回答由提问者推荐
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
