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
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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sin2x是sin平方x)在线等
![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](/uploads/image/z/17049030-6-0.jpg?t=%E8%8B%A5f%281%29%3D0.%E4%B8%94f%27%281%29%E5%AD%98%E5%9C%A8%E6%B1%82lim%28f%28sin2x%2Bcosx%29%2F%28e%5Ex-1%29tanx%29%28x+%E8%B6%8B%E5%90%91%E4%BA%8E0%2C%E5%85%B6%E4%B8%ADsin)
=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)
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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