Analysis Question, I posted yesterday!

1. Mar 23, 2008

Math_Geek

[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?
Thanks,
Michelle

2. Mar 23, 2008

HallsofIvy

Staff Emeritus
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 $|x-3|< \epsilon$ then $|f(x)- 6|= |x+3-6|= |x-3|< \epsilon$. If x is rational then |f(x)- 6|= |2x- 6|= 2|x-3|. In order to make that less than $\epsilon$ 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?

3. Mar 23, 2008

Math_Geek

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.

4. Mar 23, 2008

tiny-tim

Hi Michelle!

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!