lim→x-a (sin^2x-sin^2a)/(x-a)
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lim→x-a (sin^2x-sin^2a)/(x-a)
lim→x-a (sin^2x-sin^2a)/(x-a)
lim→x-a (sin^2x-sin^2a)/(x-a)
估计题目的极限是x→a,这样的话.原式=x-a (sin^2x-sin^2a)/(x-a)=x-a*(sinx+sina)(sinx-sina)/(x-a)<计此式为(1)式>,由于x→a,则在(1)式中sinx可以等于sina,即sinx+sina=2sina,则(1)式=x-a*2sina*(sinx-sina)/(x-a)<计为(2)式>,此处sinx-sina与x-a等价,(2)式=x-a*2sina=a-2asina
(以上两步等价可以用泰勒展开)
lim(x→a)[2cos(x+a)sin(x-a)]/(x-a)=2cos2a
lim→x-a (sin^2x-sin^2a)/(x-a)
=lim→x-a (sinx-sina)(sinx+sina)/(x-a)
=2sina*lim→x-a (sinx-sina)/(x-a)
=2sina*cosa
=cos2a
用罗比达法则 =2cos2x
lim(x(sin)^2)
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