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(2011•北京模拟)设f(x)=A.极限不存在B.极限存在,但不连续C.连续,但不可导D.可导

题目详情
(2011•北京模拟)设f(x)=
A.极限不存在
B.极限存在,但不连续
C.连续,但不可导
D.可导
▼优质解答
答案和解析
①f′(0+0)=
lim
x→0+
f(x)−f(0)
x
=
lim
x→0+
1−cosx
x
−0
x
=
lim
x→0+
1−cosx
x
3
2
=
lim
x→0+
1
2
x2
x
3
2
=
1
2
lim
x→0+
x
1
2
=0;
f′(0−0)=
lim
x→0−
f(x)−f(0)
x
=
lim
x→0−
x2g(x)−0
x
=
lim
x→0−
xg(x)=0.
即f(x)在x=0处左、右导数存在且相等,因此,f(x)在x=0处可导,故选项C错误,选项D正确;
lim
x→0+
f(x)=
lim
x→0+
1−cosx
x
=
lim
x→0+
1
2
x2
x
1
2
lim
x→0+
x
3
2
=0;
lim
x→0−
f(x)=
lim
x→0−
x2g(x)=0,函数左、右极限存在且相等,则函数f(x)在x=0处极限存在,故选项A错误;
lim
x→0+
f(x)=
lim
x→0−
f(x)=f(0)=0,则函数f(x)在x=0处连续,故选项B错误;
故选:D.
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