Question

若f(x)在x=0的某个邻域中有连续的一阶导数f’(0)=0,f”(0)存在,

Answer 1

用拉格朗日中值定理f(x)-f(sinx)=f'(t)(x-sinx),t∈(sinx,x)原极限=f'(t)(x-sinx)/x^4=f'(t)(x-x+x^3/6+o(x^4))/x^4=f'(t)(x^3/6+o(x^4))/x^4=1/6*f'(t)/x=1/6*[f'(t)-f'(0)]/(t-0) *t/x=1/6*f''(0)*t/x由于sinx...