lim[(x^2+1)/(x+1)-(ax+b)]=1其中x趋近无穷大 ,求常数a,b的值
人气：423 ℃ 时间：2020-05-19 07:51:51
解答
lim【x→∞】[(x²+1)/(x+1)-(ax+b)]
=lim【x→∞】[(x²+1)-(x+1)(ax+b)]/(x+1)
=lim【x→∞】[(1-a)x²-(a+b)x+1-b]/(x+1)
=1
因为上面的极限存在
所以1-a=0
-(a+b)=1
解得a=1
b=-2
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