若f(1)=0.且f'(1)存在求lim(f(sin2x+cosx)/(e^x-1)tanx)(x 趋向于0,其中sin

=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)

不能用洛毕塔，因为没告诉在x=1处的某去心领域可导，
答案还是f'(1)/2
正确答案见参考资料