x->0sinx~ x-(1/6)x^3x-sinx~ (1/6)x^3------lim(x->0) f(x)=lim(x->0) [1/sinx-cosx/x]=lim(x->0) [x-sinx]/(xsinx]=lim(x->0) (1/6)x^3/x^2=0=f(0)x=0, f(x)连续
