高分求解数学问题啊(99+)关于differentiable and continuous的

发布网友发布时间：2024-02-03 13:33

共5个回答

热心网友时间：2024-03-05 01:15

h(x)= {x³sin(1/x),x≠0
｛0，x=0
h'(0-)=lim(x→0-)[h(x)-h(0)]/x=lim(x→0-)h(x)/x=lim(x→0-)x²sin(1/x)=0
h'(0+)=lim(x→0+)[h(x)-h(0)]/x=lim(x→0+)h(x)/x=lim(x→0+)x²sin(1/x)=0
∴h'(0-)=h'(0+)=0
∴h(x)在点x=0处可导 (h(x) is differentiable at 0 point)
while x≠0，h'(x)=3x²sin(1/x)+x³cos(1/x)*(-1/x²)=3x²sin(1/x)-xcos(1/x)
∴h(x)在R上可导(h is differentiable everywhere )

h'=3x²sin(1/x)-xcos(1/x)
h''(0-)=lim(x→0-)[h'(x)-h'(0)]/x=lim(x→0-)[3sin(1/x)-cos(1/x)]=lim(x→0-)[√10sin(1/x -a)]极限不存在
∴h'(x)在x=0处不可导（h'(x) is differentiable at 0 point)
易知h'(x)在x≠0处可导，∴ h' is continuous everywhere but fails to have a derivative at 0 point.

热心网友时间：2024-03-05 01:16

电脑输入太慢了照片吧

热心网友时间：2024-03-05 01:15

h=0吗？也就只有这一点不可导吧，不太清楚

热心网友时间：2024-03-05 01:16

h(x)=x^3sin(1/x)处处可导
当x≠0时 h'(x)=3x^2sin(1/x)- x cos(1/x)
当x≠0时 h'(0)=lim(x→0) [x^3sin(1/x)-0]/x =limx^2sin(1/x)=0
所以显然 h'(x)在x≠0 处处连续因为都有定义是个初等函数
当x=0 limh'(x)=lim2x^2sin(1/x)- x cos(1/x) = 0=h'(0)
所以h'(x)仅处处连续
但没有一点可导因为
h''(x)= lim (x→0)[h'(x)-h'(0)]/x
lim [h'(x)-h'(0)]/x = lim 3x sin(1/x)- cos(1/x)
显然极限不存在所以处处不可导

热心网友时间：2024-03-05 01:14