Question

lim(x→+∞)[√(x²-x+1)-ax+b]=0,求a,b的值

Answer 1

lim(x->+∞)[√(x^2-x+1) -ax+b] =0
let
y=1/x

lim(x->+∞)[√(x^2-x+1) -ax+b] =0
lim(y->0)[ (1/y)[ √(y^2-y+1) -a] +b] =0

y->0
√(y^2-y+1)
=√[1+(-y+y^2)]
~ 1 + (1/2)(-y+ y^2)
~ 1 - (1/2)y

(1/y)[ √(y^2-y+1) -a] +b
~(1/y)[ (1-a) - (1/2)y ] +b
=(1/y)(1-a) +(b- 1/2)

=>
1-a =0 and b-1/2=0
a=1 and b=1/2

追问

=√[1+(-y+y^2)]
~ 1 + (1/2)(-y+ y^2)
~ 1 - (1/2)y
这个是怎么出来的

追答

在等价无穷小， 考虑 第一个出现的最低价
√(1+x) ~ 1+ (1/2)x

√[1+(-y+y^2)]
~ 1 + (1/2)(-y+ y^2)
~ 1 -(1/2)y

追问

1/2y²去哪了？

追答

y->0
lim(y->0) (1/y)[(1/2)y^2]

=lim(y->0) [(1/2)y]
=0
这个项不用考虑！

追问

1 + (1/2)(-y+ y^2)
~ 1 -(1/2)y
这步的1/2y²去哪了

追答

考虑第一个出现的最低价
1 + (1/2)(-y+ y^2)
=1 - (1/2)y+ (1/2)y^2
~ 1 -(1/2)y

(1/2)y^2 这个项不用考虑, 因为
lim(y->0) (1/y)[(1/2)y^2]
=lim(y->0) [(1/2)y]
=0