1个回答
展开全部
(3)
f(x)
=x+4 ; x<1
=2x+3 ; x≥1
f(1-)=lim(x->1-)(x+4) =5
f(1) =f(1+) =lim(x->1+)(2x+3) =5 = f(1-)
lim(x->1) f(x) =5 =f(1)
x=1, f(x) 连续
(4)
f(x)
=x+1 ; x<3
=0 ; x=3
=2x-3 ; x>3
f(3-)=lim(x->3-) (x+1) =4
f(3+) =lim(x->3+) (2x-3) = 3 ≠f(3-)
lim(x->3) 不存在
f(x)
=x+4 ; x<1
=2x+3 ; x≥1
f(1-)=lim(x->1-)(x+4) =5
f(1) =f(1+) =lim(x->1+)(2x+3) =5 = f(1-)
lim(x->1) f(x) =5 =f(1)
x=1, f(x) 连续
(4)
f(x)
=x+1 ; x<3
=0 ; x=3
=2x-3 ; x>3
f(3-)=lim(x->3-) (x+1) =4
f(3+) =lim(x->3+) (2x-3) = 3 ≠f(3-)
lim(x->3) 不存在
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询