Prove f is not piecewise continuous on [-1,1]

[maupassant]

Homework Statement

Let f(x) = x sgn(sin(1/x)) if x != 0
f(x) = 0 if x = 0
on the interval I=[-1,1]

Now I 'm asked to show that f(x) is not piecewise continuous on I and later, I must show that f is integrable on I.

The Attempt at a Solution

I am completely lost here and don't know how to proceed...
I think we should first divide the partition in some regular subpartitions so that x=0 is excluded (the only possible discontinuity?):
P1=[-1, -e/100] and P2=[e/100, 1], and then try to show that continuity on the interval P3=[-e/100, e/100] is impossible but that we still have uniform continuity on P1 and P2?
Should I be looking in that direction? Or should I look at it in a complete different way?

Thank you for the help!

 

[CompuChip]

Intuitively, it is not piecewise continuous because sin(1/x) changes sign arbitrarily often as you let x tend to zero (1/x grows without bound, so there is always an x arbitrarily close for which the sign of the sine has flipped).

So maybe you can use this for a rigorous proof. For example, suppose that sin(1/x) is piecewise continuous. Then you can list all points of discontinuity and there are finitely many, for example x₁ < x₂ < ... < x_n.
If you can show that there must be some 0 < x₀ < x₁, such that sin(1/x₀) has a sign opposite from sin(1/x₁), you are done (you have listed them all, and found another one, which is a contradiction).

 

[lanedance]

I assume you mean the sign function
http://en.wikipedia.org/wiki/Sign_function

your function will be discontinous everytime sin(1/x) = 0, or when 1/x = n.pi,

so when x = 1/(n.pi)

if you definition for piecewise continuous contains discontinuous at only finitely many discontinuities, it sould be easy to show this is contradicted

 

[lanedance]

right, so just like Compuchip said ;)

 

[maupassant]

Thank you very much for your help guys!