=lim[f(sin^2x+cosx)/x^2]（等价无穷小的替换）因f'(1)存在则f(1+x)-f(1)=f'(1)x+o(x)即f[1+(sin^2x+cosx-1)]=f'(1)(sin^2x+cosx-1)+o(sin^2x+cosx-1)=f'(1)(sin^2x+cosx-1)+o(x^2)从而lim[f(sin^2x+cosx)/x^2]=lim{[f'(1)(sin^2x+cosx-1)+o(x^2)]/x^2}=f'(1)lim[(sin^2x+cosx-1)/x^2]=f'(1)lim[(2sinxcosx-sinx)/(2x)]=1/2*f'(1)lim(2cosx-1)*lim(sinx/x)=1/2*f'(1)
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