暗影寂
幼苗
共回答了20个问题采纳率:95% 举报
解.当x→0时,x²-2x-3是无穷小量,要使极限存在,则分母也是无穷小量即
9+3a+b=0即b=-3a-9
lim(x²+ax+b)/(x²-2x-3)=lim(x²+ax-3a-9)/(x²-2x-3)
=lim[(x+3)(x-3)+a(x-3)]/[(x+1)(x-3)]
=lim[(x-3)(x+3+a)]/[(x+1)(x-3)]
=lim(x+3+a)/(x+1)
=(3+3+a)/(3+1)=5
解得a=14
b=-3*14-9=-51
1年前
3