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Thread: Rationals and Irrationals

  1. #1
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    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 $\displaystyle (q_n) = c + \frac{1}{n}$ 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!
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  2. #2
    Member mabruka's Avatar
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    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??
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  3. #3
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    Sorry that's my fault, $\displaystyle c$ is a cluster point of the function $\displaystyle f(x)$
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  4. #4
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    The problem is to show that "if $\displaystyle c\ne 0$ then $\displaystyle \lim_{x\to c} f(x)$ 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 $\displaystyle \lim_{x\to 0} f(x)= L$. Take $\displaystyle \epsilon= \frac{|L|}{2}$ and use the fact that every interval contains both rational and irrational numbers.
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