设f(0)=0 f ' (0)=3 求lim f(tanx-sinx)/x^2ln(1-x) 其中x趋近于0

ln(1-x)等价于-x
lim[f(0+tanx-sinx)-f(0)]/(tanx-sinx)=f'(0)=3
lim(tanx-sinx)/x^3=limtanx(1-cosx)/x^3=lim(x*x^2/2)/x^3=1/2
所以原式=lim{[f(0+tanx-sinx)-f(0)]/(tanx-sinx)}*{(tanx-sinx}/(-x^3)=-f'(0)/2=-3/2