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Homework Help: Analysis Question, I posted yesterday!

  1. Mar 23, 2008 #1
    [SOLVED] Analysis Question, I posted yesterday!

    1. The problem statement, all variables and given/known data
    Define f(x)=2x, x is rational and x+3 when x is irrational. Find all points where g(x) is continuous and prove continuity at these points

    2. Relevant equations
    From analysis homework and using the real definition of continuity

    3. The attempt at a solution
    what I did was I took the limit as x approached 3 of x+3, and found for 0<|x+3|<epsilon which will equal delta, then |X+3-6|=|x-3| which is less than epsilon. So then I know that the limit of the function =the limit of f(3) so therefore continuous at x=3. Is there anymore that I need?
  2. jcsd
  3. Mar 23, 2008 #2


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    Any more? What you've done is not correct because the function you are dealing with is not x+3! You say nothing about why you are interested in
    x= 3; could it possibly be because you have decided the function is only continuous at x= 3? It would have been a good idea to start by saying that and saying why you think that. As long as x is irrational and [itex]|x-3|< \epsilon[/itex] then [itex]|f(x)- 6|= |x+3-6|= |x-3|< \epsilon[/itex]. If x is rational then |f(x)- 6|= |2x- 6|= 2|x-3|. In order to make that less than [itex]\epsilon[/itex] how small must |x-3| be? How can you make sure both of those are true.

    And what about if x is not equal to 3? If f were continuous at x= a then the limit would be f(a). That, in turn, would mean that the limit of any sequence {f(xn)}, where xn converges to a, would have to be f(a). Suppose xn were a sequence of rational numbers converging to a- what would {f(xn)} converge to? Suppose xn were a sequence of irrational numbers converging to a- what would {f(xn)} converge to? What does that tell you?
  4. Mar 23, 2008 #3
    Both of these lines will intersect at the point (3,6), and you would have to make epsilon= e/2 to satisfy both functions when x is either rational or irrational.
  5. Mar 23, 2008 #4


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    Hi Michelle! :smile:

    This very unclear …

    For a start, you meant 0<|x-3|<epsilon.

    And I know you meant "in this case we can choose our delta equal to epsilon", but there are better ways of saying that.
    hmm … a bit late …

    Try writing it out, all in one go, and clearly! :smile:
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