Question

求详细解答过程及思路

Answer 1

解：
lim【x→∞】ax²/(x+1)+bx
=lim【x→∞】(ax²+bx²+bx)/(x+1)
=lim【x→∞】[(a+b)x²+bx]/(x+1)
=lim【x→∞】[(a+b)x+b]/(1+1/x)
因为该极限为2
所以a+b=0

答案：a+b=0

Answer 2

lim( ( ax^2+bx^2 +bx ) / (x+1) )=lim[(a+b)x^2] / (x+1) +lim bx /(x+1)
当X趋向正无穷的时候，x^2 /（x+1）趋向于正无穷，所以x^2的系数只能是0
所以 a+b=0
lim bx /(x+1)=2 所以 b=2