利用等价无穷小的性质计算：lim(x→0)时,tan(2x^2)/1-cosx
展开全部
(tan2x)^2=sin2x/cos2x=4sin(x/2)cos(x/2)cosx/cos2x
1-cosx=2[sin(x/2)]^2
(tan2x)^2/(1-cosx)=2cos(x/2)cosx/[cos2xsin(x/2)]=[cos(3x/2)+cos(x/2)]/[cos2xsin(x/2)]
lim(x→0)(tan2x)^2/(1-cosx)=lim(x→0) [cos(3x/2)+cos(x/2)]/[cos2xsin(x/2)]
(x→0),sin(x/2)→0,cos3x/2 →1 cosx/2 →1
lim(x→0)(tan2x)^/(1-cosx)=∞
1-cosx=2[sin(x/2)]^2
(tan2x)^2/(1-cosx)=2cos(x/2)cosx/[cos2xsin(x/2)]=[cos(3x/2)+cos(x/2)]/[cos2xsin(x/2)]
lim(x→0)(tan2x)^2/(1-cosx)=lim(x→0) [cos(3x/2)+cos(x/2)]/[cos2xsin(x/2)]
(x→0),sin(x/2)→0,cos3x/2 →1 cosx/2 →1
lim(x→0)(tan2x)^/(1-cosx)=∞
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
