求limx→0(根号2-根号(1+cosx))/(sin3x)^2的极限

求limx→0(根号2-根号(1+cosx))/(sin3x)^2的极限
求详解
feihuoxiaozi 1年前 已收到1个回答 举报

lwzyhm 幼苗

共回答了21个问题采纳率:81% 举报

limx→0(√2-√(1+cosx))/(sin3x)^2=lim(1-cosx)/[(sin3x)^2(√2+√(1+cosx))]
=lim(1-(1-2sin^2(x/2)))/[(sin3x)^2(√2+√(1+cosx))]
=lim2sin^2(x/2)/[(sin3x)^2(√2+√(1+cosx))]
=lim2sin^2(x/2)*1/4/(x/2)^2/[(sin3x)^2*9/(3x)^2*(√2+√(1+cosx))]
=2*1/4/9/2√2=√2/72

1年前

4
可能相似的问题
Copyright © 2024 YULUCN.COM - 雨露学习互助 - 16 q. 0.028 s. - webmaster@yulucn.com