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讨论f(x)=
{x
2
-1,0≤x<1,
x+1,x>1
在x=1点的连续性.
分类:
高等数学(一)(00020)
发表:2024年09月10日 02时09分19秒
作者:
admin
阅读:
(19)
讨论f(x)=
{x
2
-1,0≤x<1,
x+1,x>1
在x=1点的连续性.
【正确答案】:所给函数是个分段函数,x=1点是f(x)的分段点,f(x)在x=1点处及其附近有定义, 但是lim
x→1+
f(x)=lim
x→1+
(x+1)=2,lim
x→1-
f(x) =lim
x→1
lim(x
2
-1)=0,lim
x→1-
f(x)≠lim
x→1-
f(x), 故f(x)在x=1点处无极限,从而f(x)在x=1点处不连续.
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