Let f(x)=(x−1)(x4+ax3+bx2+cx+d)
f(x)+f′(x)+f′′(x)=x5+(a+4)x4+(b+3a+16)x3+
(9a+2b+c−12)x2+(d+c+4b−6a)x+(c−2b)
∵f(x)+f′(x)+f′′(x)=x5+64
⇒a+4=0,b+3a+16=0,9a+2b+c−12=0,d+c+4b−6a=0
⇒a=−4,b=−4,c=56,d=−64
So x→1limx−1f(x)=x→1lim(x4−4x3−4x2+56x−64)
=1−4−4+56−64=−15