Differentiability

i think u know how to do limit right so i did not show the step of limit
and it is hard to show u here how to do it......
and u did not say which way does it go for xsin(1/x)

limit x go to 0 for xsin(1/x)=0

xcos(1/x)*-(x^-2)+sin(1/x)
F'(x)=
0

limit x go to +0 for xcos(1/x)*-(x^-2)+sin(1/x)= -infinitely
or
limit x go to -0 for xcos(1/x)*-(x^-2)+sin(1/x)= +infinitely

-infinitely
therefore the F'(0) is not exist as x at 0 the F'(0)=
0

2006-10-19 01:01:57 補充：
....the yahoo show F'(x) is = dF(x)/dxsorry about that so _________| xcos(1/x)*-(x^-2)+sin(1/x)dF(x)/dx= _________|............... 0

2006-10-20 18:25:39 補充：
mathmouse:seen u look like a university student u should know this concept.For a function y=f(x) is differentiable at x=a if:1) y=f(x) is continuous at x=a2) lim (x go to a from the left) df(x)/dx = lim (x go to a from the right) df(x)/dx

2006-10-20 18:27:59 補充：
Form point 1 we know that for a function y=f(x) is continuous at the point x=a if:1) f(a) must exist2) lim (x go to a from the left) f(x) = lim (x go to a from the left) f(x)=f(a)

2006-10-20 18:29:03 補充：
therefore1) f(0) = 02) lim (x go to 0 from the left) xsin(1/x)=lim (x go to 0 from the left) xsin(1/x)=f(0)3) lim (x go to 0 from the left) df(x)/dx=lim (x go to 0 from the right) df(x)/dx　　　　　╭　　　　　l　xcos(1/x)*-(x^-2)+sin(1/x) = -(1/x)cos(1/x)+sin(1/x)df(x)/dx =　l　　　　　l　　　　　　0　　　　　╰

2006-10-20 18:30:19 補充：
so,lim (x go to 0 from the left) df(x)/dx = infinitelylim (x go to 0 from the right) df(x)/dx = - infinitelybecause lim (x go to a from the left) df(0)/dx is not equallim (x go to a from the right) df(0)/dx therefore it is not differentiable at x= 0

2006-10-22 18:18:51 補充：
mathmouse:*wrong concept!What you have shown is the discontinuity of dF/dx at x = 0**lim(x －﹥a from left)df(x)/dx=lim(x －﹥a from right)df(x)/dx*since u not really understand differentiability so just want to explain why we can use the above statement to define is it differentiable at a point

2006-10-22 18:20:54 補充：
for a point where differentiability to be exist the function of df(x)/dx at that point must be continuousfor example (one where is not differentiable):y=|x|　at x=0for this function we all know that it cannot differentiable at x=0 as it has a shape end point although it is continuous at x=0

2006-10-22 18:22:21 補充：
due to the shape end point so we don't know the rate of change at x=0remember the proof of the first principles of differentiation this will explain whyso,　　　╭　　 　 |　1　x＞0dy/dx =|　 　 　|　-1　x＜0　　　╰

2006-10-22 18:22:33 補充：
which lim(x －﹥a from the left) df(x)/dx≠lim(x －﹥a from the left) df(x)/dx ∴ it is not differentiable at x=0

A formal proof:
Consider the limit lim[x -> 0] [(F(x) - F(0)) / (x - 0)]. If this limit exist, then F is differentiable at x = 0.
Note that
(F(x) - F(0)) / (x - 0)
= [xsin(1 / x) - 0] / x
= sin(1 / x)
However, lim[x -> 0] sin(1 / x) does not exist.
Therefore F is NOT differentiable at x = 0.


2006-10-19 06:49:31 補充：
To show that lim[x -> 0] sin(1 / x) does not exist, you can consider two convergent sequences:x_n = 1 / 2nπ, y_n = 1 / (2n + 1)πNote that sin(1 / x_n) = sin 2nπ -> 0 but sin(1 / y_n) = sin(2n + 1)π -> 1.

2006-10-19 06:53:09 補充：
choi_mickey: wrong concept! What you have shown is the discontinuity of dF / dx at x = 0 (if it exists). myisland8132: It's just an intuitive proof. It's a good idea to consider the graph. But to verify the guess, you need to prove it rigorously.

　　　　╭
　　　　　　　　　　　 l　xsin(1/x)　(x no equal 0)
Determine if 　　　F(x)= l　　　　　　　　　　　　　is differentiable x=0
　　　　　　　　　　　 l　0　　　　( x equal 0 )

it can not differentiate at x=0.
From the graph of F(x), we see that when the point tends to 0, the secant line y
oscillatiates between y=x and y=-x, so F(x) cannot differentiate

2006-10-19 01:37:35 補充：
問題是問可不可以 differentiable 不是問是否在x=0連續﹐若問是否連續﹐則答案是對的Prove that lim x- 0 of x sin(1/x) = 0 |xsin(1/x)| = |x||sin(1/x)| 0 of both -|x| and |x| = 0 Thus by pinching theorem lim x- 0 xsin(1/x) = 0

2006-10-19 14:43:51 補充：
Suppose the limit did exist, then there would be an L such that given an ε＞0,then |x| ＜ σ such that|sin(1/x)-L| ＜εlet x1=1/(Nπ) such that |x1| ＜ σ|sin(1/x1)-L|=|L|＜ε

2006-10-19 14:45:15 補充：
now let x2=1/[(2n 1/2)π] such that |x2| ＜ σ|sin(1/x2)-L|=|1-L|＜εbut if take ε＜1/2, L will be close to both 0 and 1,a contradiction.lim(x→0)(sin1/x) does not exist

2006-10-19 14:49:34 補充：
mathmouse:因為以前看個本大學教科書無寫到個formal proof﹐以為好難證(指內容上)﹐不過你講完用1/(Nπ)和1/[(2n 1/2)π]﹐原來都係大學生可以做到的程度﹐不過係技巧問題。