第十二题咋写啊

分子
f(x)=ax^4+bx^3+1 =>f(1)=a+b+1
f'(x)=4ax^3+3bx^2 =>f'(1)/1! = 4a+3b
f''(x)=12ax^2+6b =>f''(1)/2! = 6a+3b
ie
ax^4+bx^3+1 =(a+b+1)+(4a+3b)(x-1)+(6a+3b)(x-1)^2 +o[(x-1)^2]
分母
sinπx
=sin(π-πx)
=sin[π(1-x)]
=-π(x-1) +o(x-1)
(x-1)sinπx =-π(x-1)^2 +[o(x-1)^2]
lim(x->1) (ax^4+bx^3+1)/[(x-1)sinπx]=c
=>
a+b+1=0 (1)
4a+3b =0 (2)
-(6a+3b)/π = c (3)
4(1) -(2)
b=-4
from (1)
a+b+1=0
a-4+1=0
a=3
from (3)
-(6a+3b)/π = c
c=-(18-12)/π =-6/π