是不是这样

x->0
tanx = x +(1/3)x^3 +o(x^3)
xf(x) = xf(0) +x^2.f'(0) +(1/2)f''(0)x^3 +o(x^3)
tanx +xf(x) = [1+f(0)]x + f'(0)x^2 + [ 1/3 + (1/2)f''(0)]x^3 +o(x^3)
lim(x->0) [tanx +xf(x)]/x^3 =0
=>
1+f(0)=0 and f'(0)=0 and 1/3 + (1/2)f''(0) =0
f(0)=-1 and f'(0)=0 and (1/2)f''(0) = -1/3
lim(x->0) [1+f(x)]/x^2
=lim(x->0) { [1+f(0)]x + f'(0).x + (1/2)f''(0)x^2 }/x^2
=lim(x->0) [ -(1/3)x^2 ]/x^2
=-1/3

把tanx展开为x+x³/3+o(x³)，带入得1/3+所求极限=0

感谢感谢