已知函数f(x)=(x+b)ex(x<0)x3+2a(x≥0)(a≠0)在点x=0处连续,则limx→∞[1x2-x-ba(x2-2x)]=()A.-1B.0C.-12D.1-数学

题目简介

已知函数f(x)=(x+b)ex(x<0)x3+2a(x≥0)(a≠0)在点x=0处连续,则limx→∞[1x2-x-ba(x2-2x)]=()A.-1B.0C.-12D.1-数学

题目详情

已知函数f(x)=
(x+b)ex(x<0)
x3+2a(x≥0)
(a≠0)
在点x=0处连续,则
lim
x→∞
[
1
x2-x
-
b
a(x2-2x)
]
=(  )
A.-1B.0C.-
1
2
D.1
题型:单选题难度:中档来源:广西一模

答案

∵已知函数f(x)=
(x+b)ex(x<0)
x3+2a(x≥0)
(a≠0)
在点x=0处连续,
∴b×1=2a,
lim
x→∞
[class="stub"1
x2-x
-class="stub"b
a(x2-2x)
]
=
lim
x→∞
[class="stub"1
x2-x
-class="stub"2
(x2-2x)
]
=
lim
x→∞
class="stub"-x
x(x-1)(x-2)
=
lim
x→∞
class="stub"-1
(x-1)(x-2)
=0,
故选B.

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