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 $\displaystyle (q_n) = c + \frac{1}{n}$ for all $\displaystyle n$ in the naturals, which is a sequence of rationals. Then I let $\displaystyle (i_n) = c + \frac{\sqrt{2}}{n}$ 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 $\displaystyle (i_n) = c + \frac{1}{n}$ 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!