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Homework Help: Continuety of these functions

  1. Dec 27, 2008 #1
    describe the continuety of these functions
    http://img233.imageshack.us/img233/6141/76604020ql2.gif [Broken]

    regarding the [] functions i learned this fact

    http://img184.imageshack.us/img184/2469/87823960yl2.gif [Broken]
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 27, 2008 #2


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    Since you used "[x]" I was not sure if you meant "floor of x", [itex]\floor(c)[/itex] or "ceiling of x", [tex]\ceil(x)[/itex]. From the I'm second link it appears that you mean the "floor" function.
    Yes, [itex]\floor(x)[/itex] is constant between intgers: n-1 from n-1 up to n and n from n to n+1 so it is really only necessary to look at what happens as you approach the integer n from below and above.

    The last one appears to be "piecewise", defined by different formulas for x rational and irrational but there appears to be something written before "[itex]\pi x[/itex] for x rational. Is that "sin"?
    Last edited by a moderator: Dec 29, 2008
  4. Dec 28, 2008 #3
    yes its sin

    does it change your answer?
  5. Dec 29, 2008 #4
    how to do a limit of this x*1/[x]

    i get 0*+infinity
  6. Dec 29, 2008 #5
    how to solve the splitted
    i cant imagine the graph
    i cant do a limit
    because rational and irrational are infinitely mixed
    what to do?
  7. Dec 29, 2008 #6


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    For x any number other than an integer, if x is between n and n+ 1, [x]= n so
    x/[x]= x/n and x[x]= xn. It should be obvious that the function is continuous there. For x slightly below n, say x= n-y, x/[x]= n-y/(n-1) and x[x]= (n-y)n-1= xn/xy. For x slightly larger than n, say x= n+ y, x/[x]= (n+y)/n and x[x]= (n+y)n. Are the limits of those, as y goes to 0, the same?

    As for [itex]sin(\pi x)[/itex] for x rational, 0 for x irrational, use the fact that [itex]\lim_{x\rightarrow a} f(x)= L[/itex] if and only if [itex]\lim_{n\rightarrow \infty} f(a_n)= L[/itex] for any sequence [itex]{a_n}[/itex] converging to x. In particular, for any number x, there exist a sequence of irrational numbers converging to x so for such a sequence the limit of this sequence will be 0. That means that in order that the limit itself exist at x, we must also have [itex]sin(a_n\pi)[/itex] converge to 0 also. Since sine is a continuous function, it is easy to see that that limit is [itex]sin(\pi x)[/itex]. That is, this function is continuous exactly for those x such that [itex]sin(\pi x)= 0[/itex].
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