利用等价无穷小的性质计算：lim(x→0)时,tan(2x^2)/1-cosx
问题描述：
利用等价无穷小的性质计算：lim(x→0)时,tan(2x^2)/1-cosx
答
(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)=∞