发布网友 发布时间:2024-10-23 20:44
共1个回答
热心网友 时间:2024-11-01 15:49
令f(t)=cost,g(t)=(t^2)/3+1
根据柯西中值定理,存在k介于x与2x之间,使得:f'(k)/g'(k)=[f(x)-f(2x)]/[g(x)-g(2x)]
(-sink)/(2k/3)=(cosx-cos2x)/[(x^2)/3-(4x^2)/3]
(3/2)*(sink)/k=(cosx-cos2x)/(x^2)
所以lim(x->0) (cosx-cos2x)/(x^2)
=lim(k->0) (3/2)*(sink)/k
=3/2