求极限x趋于0时(1-cos(x/2))x / (tan x-sin x)关键是答案中有一步化简看不懂cos x*x*(1-cos(x/2)) / sinx(1-cos x)=1-cos(x/2) / 1-cos x

来源:学生作业帮助网 编辑:作业帮 时间:2024/03/30 00:01:49
求极限x趋于0时(1-cos(x/2))x / (tan x-sin x)关键是答案中有一步化简看不懂cos x*x*(1-cos(x/2))  /  sinx(1-cos x)=1-cos(x/2)  /  1-cos x

求极限x趋于0时(1-cos(x/2))x / (tan x-sin x)关键是答案中有一步化简看不懂cos x*x*(1-cos(x/2)) / sinx(1-cos x)=1-cos(x/2) / 1-cos x
求极限x趋于0时(1-cos(x/2))x / (tan x-sin x)
关键是答案中有一步化简看不懂
cos x*x*(1-cos(x/2)) / sinx(1-cos x)
=1-cos(x/2) / 1-cos x

求极限x趋于0时(1-cos(x/2))x / (tan x-sin x)关键是答案中有一步化简看不懂cos x*x*(1-cos(x/2)) / sinx(1-cos x)=1-cos(x/2) / 1-cos x
答案见图片

lim(x->0)[(1-cos(x/2))x] '/ [(tanx-sinx)]'
=lim(x->0) [(1/2)sin(x/2)x+(1-cos(x/2)]/(1/cosx^2-cosx)
=lim(x->0)[ [(1/2)sin(x/2)x+(1-cos(x/2)]cosx^2 ]' / [ (1-cosx^3) ]'
=lim(x->0)[(1/4)cos...

全部展开

lim(x->0)[(1-cos(x/2))x] '/ [(tanx-sinx)]'
=lim(x->0) [(1/2)sin(x/2)x+(1-cos(x/2)]/(1/cosx^2-cosx)
=lim(x->0)[ [(1/2)sin(x/2)x+(1-cos(x/2)]cosx^2 ]' / [ (1-cosx^3) ]'
=lim(x->0)[(1/4)cos(x/2)xcosx^2 +[(1/2)sin(x/2)x+(1-cos(x/2)](-2sinxcosx) ] /[-3cosx^2]
=0

收起

对于cos x*x*(1-cos(x/2)) / sinx(1-cos x)
x趋于0时,cosx趋于1,x/sinx趋于1
所以=1-cos(x/2) / 1-cos x
=2sin²(x/4)/2sin²(x/2)
={sin²(x/4)/(x/4)...

全部展开

对于cos x*x*(1-cos(x/2)) / sinx(1-cos x)
x趋于0时,cosx趋于1,x/sinx趋于1
所以=1-cos(x/2) / 1-cos x
=2sin²(x/4)/2sin²(x/2)
={sin²(x/4)/(x/4)² }*{(x/2)²/sin²(x/2)} *1/4
—›1/4 (x—›0)
关键是lim(x->0) sinx/x=1的运用

收起

1/4