limx趋近于0ln（1-2x）+2xf（x）？

x->0ln(1-2x) = (-2x) -(1/2)(-2x)^2 +o(x^2)=-2x - 2x^2 +o(x^2)lim(x->0) [ln(1-2x)+2xf(x) ]/x^2 =0=>x->02xf(x) =-[-2x - 2x^2 +o(x^2)] f(x) = 1+x +o(x)f'(0) = 1

匿名回答于2024-06-06 10:02:48