因为lim[x→1] (x-1)=0
又lim[(x-1)/(2x^3+ax+b)]=1/4
所以2x^3+ax+b= (x-1)(kx^2+cx+e)= kx^3+cx^2+ex-kx^2-cx-e =2x^3+ax+b
对比系数得到:k=c=2 ,e-c=e-2=a ,b=-e
且又有lim[x→1](kx^2+cx+e) = k+c+e=2+2+e=4 ,e=0 ,b=0,a=-2