piece wise continuity

barksdalemc

I have a question that asks to show that f(x)=x^2(sin[1/x]) is piecewise continuous in the interval (0,1). I need to show that I partition the interval in to finite intervals and the function is continous within the subintervals and have discontinuities of te first type at the endpoints. I tried using any multiple of pi. That doesn't work. Any hints?

StatusX

Is that:

[tex]x^2 \sin \left \frac{1}{x} \right [/tex]

? Because that function is just continuous in (0,1), so it's obviously piecewise continuous. Or does [1/x] mean the integer part of 1/x? If so, then it seems like the natural thing to do is divide up (0,1) into the regions where [1/x] takes distinct values (ie, (1/2,1],(1/3,1/2], etc.).

barksdalemc

[tex]x^2 \sin \left \frac{1}{x} \right [/tex]

Is the correct function, which you indicated. The exact question is as follows:

A function f is called piecewise continuous (sectionally continuous) on an interval (a,b) if there are finitely many point a = x(sub o) < x(sub 1) < ... < x(sub n) = b such that
(a) f is continuous on each subinterval x(sub o) < x < x(sub 1) , x(sub 1) < x < x(sub 2) , ... , x(sub n-1) < x < x(sub n), and
(b) f has discontinuities of the first kind at the point x(sub o),x(sub 1), ... , x (sub n). The function f(x) need not be defined at the points x(sub o),x(sub 1), ... , x (sub n). Show that the following functions are piecewise continuous:

f(x) = [tex]x^2 \sin \left \frac{1}{x} \right [/tex] , 0 < x < 1

StatusX

Right, and do you see why that is continuous in the ordinary sense (or piecewise continuous with a trivial partition of (0,1))?

barksdalemc

It seems like the question wants a further partition of (0,1). The left limit doesn't exist here so I'm confused as to why the function has a discontinutiy of type 1 on the trivial interval (0,1)

StatusX

Well the left limit does exist, but this doesn't matter because 0 isn't in your range. The fact is there are no discontinuities anywhere. A function doesn't have to have any discontinuities to be piecewise continuous. The definition you gave above is a little awkward, but it doesn't rule out the possibility that the number of discontinuities is zero (which is finite), and this is certainly allowed.

barksdalemc

Maybe I don't understand the definition of the limit then. As we approach 0 the lim sup from the right is not the same as the lim inf from the right, therefore the limit from the right does not exist.

StatusX

What do you get as the limsup and liminf?

barksdalemc

0 for both. My bad.