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Analysis Question, I posted yesterday!

  • Thread starter Math_Geek
  • Start date
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[SOLVED] Analysis Question, I posted yesterday!

1. Homework Statement
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. Homework 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
 

Answers and Replies

HallsofIvy
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Homework Helper
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1. Homework Statement
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. Homework 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
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?
 
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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.
 
tiny-tim
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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?
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.
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.
hmm … a bit late …

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

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