求极限limX->0根号下(1-
原题应该是这样的
limx->0 √[1-(cosx)^2]/(1-cosx)???
=limx->0:√(sinx)^2/(1-cosx)
=limx->0:|sinx|/(1-cosx)
=limx->0:|2sin(x/2)cos(x/2)|/{2[sin(x/2)]^2}
=limx->0:|sin(x/2)|*cos(x/2)/[sin(x/2)]^2
=limx->0:+'-cos(x/2)/sin(x/2)
=+∞或者-∞
如果确实是cosx^2,那么
√(1-cosx^2)=|2sin(x^2/2)|除以1-cosx=2[sin(x/2)]^2,得到
|sin(x^2/2)...全部
原题应该是这样的
limx->0 √[1-(cosx)^2]/(1-cosx)???
=limx->0:√(sinx)^2/(1-cosx)
=limx->0:|sinx|/(1-cosx)
=limx->0:|2sin(x/2)cos(x/2)|/{2[sin(x/2)]^2}
=limx->0:|sin(x/2)|*cos(x/2)/[sin(x/2)]^2
=limx->0:+'-cos(x/2)/sin(x/2)
=+∞或者-∞
如果确实是cosx^2,那么
√(1-cosx^2)=|2sin(x^2/2)|除以1-cosx=2[sin(x/2)]^2,得到
|sin(x^2/2)/(x^2/2|/[sin(x/2)/(x/2]^2*(1/2)
->1/1^2*(1/2)=1/2。
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