lim(x趋向于0)(1-cosx)/x^2
人气:119 ℃ 时间:2020-03-24 04:13:00
解答
倍角公式:cosx=1-2[sin(x/2)]^2
故1-cosx=2[sin(x/2)]^2
于是
lim x->0 (1-cosx)/x^2
=lim x->0 2[sin(x/2)]^2/x^2
=lim x->0 2[(sin(x/2)/(x/2)]^2*1/4
=lim x->0 1/2*[(sin(x/2)/(x/2)]^2
=1/2
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