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
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![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](/uploads/image/z/3710642-50-2.jpg?t=%E6%B1%82%E6%9E%81%E9%99%90x%E8%B6%8B%E4%BA%8E0%E6%97%B6%281-cos%28x%2F2%29%29x+%2F+%28tan+x-sin+x%29%E5%85%B3%E9%94%AE%E6%98%AF%E7%AD%94%E6%A1%88%E4%B8%AD%E6%9C%89%E4%B8%80%E6%AD%A5%E5%8C%96%E7%AE%80%E7%9C%8B%E4%B8%8D%E6%87%82cos+x%2Ax%2A%281-cos%28x%2F2%29%29++%2F++sinx%281-cos+x%29%3D1-cos%28x%2F2%29++%2F++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...
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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
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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)...
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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
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1/4