lim(x趋向0+)x/[(1-cosx)^(1/2)]
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lim(x趋向0+)x/[(1-cosx)^(1/2)]lim(x趋向0+)x/[(1-cosx)^(1/2)]lim(x趋向0+)x/[(1-cosx)^(1/2)][(1-cosx)^(1/2)]
lim(x趋向0+)x/[(1-cosx)^(1/2)]
lim(x趋向0+)x/[(1-cosx)^(1/2)]
lim(x趋向0+)x/[(1-cosx)^(1/2)]
[(1-cosx)^(1/2)]=2sin²(x/2),
则原式可化为x/√2sin(x/2),利用等价无穷小的概念,xx趋向0+)时,sin(x/2)~x/2
则原式为√2
lim(x趋向0+)x/[(1-cosx)^(1/2)]
=lim(x趋向0+)x(1+cosx)^(1/2)/[(1-cos^2x)^(1/2)]
=lim(x趋向0+)(1+cosx)^(1/2)[x/sinx]
=√2
你学过泰勒展开么，翻翻书就明白了。（即使还没学到，也建议你翻一翻）没学过的话（1-cosx）/x^2 求极限，用洛必达法则后面一个因为tanx~x，把x换
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