1. ## Rationals and Irrationals

So f(x) is defined as f(x) = x if x is rational, and 0 if x is irrational. Using a sequential argument I need to show that if c=/=0, then f does not have a limit.

I am making 2 cases.
1) c is a rational number.
Then I can make two sequences. One sequence of rationals and another of irrationals. I let $(q_n) = c + \frac{1}{n}$ for all $n$ in the naturals, which is a sequence of rationals. Then I let $(i_n) = c + \frac{\sqrt{2}}{n}$ which is a sequence of irrationals.

My idea is to show that since the two sequences have different limit values at c, then f(x) can't have a limit at c.

2)c is an irrational number. Here to make a sequence of irrational numbers, I let $(i_n) = c + \frac{1}{n}$ but I cannot figure out how to create a sequence of rational numbers in the same fashion. Can anyone help me find this??

And also is this a correct approach to the problem. Thanks!

2. Hii

Sorry but i dont understand the problem:

So f(x) is defined as f(x) = x if x is rational, and 0 if x is irrational. Using a sequential argument I need to show that if c=/=0, then f does not have a limit.

"f does not have a limit" in which sense? What i am asking is what is the relation between that c and the function f??

3. Sorry that's my fault, $c$ is a cluster point of the function $f(x)$

4. The problem is to show that "if $c\ne 0$ then $\lim_{x\to c} f(x)$ does not exist".

(Since the only values of f(x) are 0 and 1, it does not have cluster points.)

Let c be a non-zero number and suppose $\lim_{x\to 0} f(x)= L$. Take $\epsilon= \frac{|L|}{2}$ and use the fact that every interval contains both rational and irrational numbers.