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Piece wise continuity

  1. Nov 1, 2006 #1
    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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  3. Nov 1, 2006 #2

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    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.).
     
  4. Nov 1, 2006 #3
    [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
     
  5. Nov 1, 2006 #4

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    Right, and do you see why that is continuous in the ordinary sense (or piecewise continuous with a trivial partition of (0,1))?
     
  6. Nov 1, 2006 #5
    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)
     
  7. Nov 1, 2006 #6

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    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.
     
  8. Nov 1, 2006 #7
    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.
     
  9. Nov 1, 2006 #8

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    What do you get as the limsup and liminf?
     
  10. Nov 1, 2006 #9
    0 for both. My bad.
     
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