lim{[(x＾2)+1]/[x+1]}-(ax+b)=0,求,a,x→∞

Question

lim{[(x＾2)+1]/[x+1]}-(ax+b)=0,求,a,x→∞

Answer 1

lim{[(x＾2)+1]/[x+1]}-(ax+b)
=lim{[(x＾2)+1-(ax+b)(x+1)]/(x+1)}
=lim{[(1-a)(x＾2)-(a+b)x+(1-b)]/(x+1)}
上下除以x
=lim{[(1-a)x-(a+b)+(1-b)/x]/[1+(1/x)]}
因为x→∞,所以1/x→0,即分母的极限为1
而要令lim{[(x＾2)+1]/[x+1]}-(ax+b)=0,即lim{[(1-a)x-(a+b)+(1-b)/x]/[1+(1/x)]}=0.
所以分子(1-a)x-(a+b)+(1-b)/x=0,
则1-a=0,a+b=0,
即a=1,b=-1
这就是完整的解法,若对此还有不懂的可再补充提问..