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## piece wise continuity

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?

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Recognitions:      Homework  Help  Is that:  $$x^2 \sin \left \frac{1}{x} \right$$  ? 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.).

$$x^2 \sin \left \frac{1}{x} \right$$  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) = $$x^2 \sin \left \frac{1}{x} \right$$ , 0 < x < 1

Recognitions:
Homework Help

## piece wise continuity

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

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)

Recognitions:      Homework  Help  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.

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.

Recognitions:      Homework  Help  What do you get as the limsup and liminf?

0 for both. My bad.

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