求极限 lim（cosx—cos3x)/x^2 其中x趋向于0.我只算出来是2,

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cos3x = cosx cos2x - sinx sin2x
cosx - cos3x = cosx ( 1- cos2x) + sinx sin2x = cosx * 2 sin²x + sinx sin2x
lim(x->0) (cosx - cos3x) / x²
= lim(x->0) [ cosx * 2 sin²x + sinx sin2x ] / x²
= 2 + 2 = 4
======以下答案可供参考======
供参考答案1:
应该是4，用两次罗必塔法则
lim（cosx—cos3x)/x^2=lim（-sinx+3sin3x）/2x = lim(-cosx+9cos3x)/2=(9-1)/2=4