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    Existence of a Limit

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    Let f(x) = x, if x is a rational number, and f(x) = x^2 if x is an irrational number.
    For what values of a, if any, does lim(f(x)) as x --> a exist? Justify your answer.

    I know that the answer is 0 and 1, but why? Please explain. Thank you.

    © BrainMass Inc. brainmass.com March 4, 2021, 8:09 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/proof-involving-existence-limit-piecewise-function-149000

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    Simply put, the answer is because x = 0 and x = 1 are the only numbers for which x = x^2.

    Given any two distinct rational numbers, there will always be an irrational number in between them. In math language, this is stated as "the set of rational numbers is dense in the real numbers". This means that if you choose two rational numbers, p and q such that p < q, there exists an irrational number, x, such that p < ...

    Solution Summary

    The expert examines proof involving existence of limit for piecewise functions. Rational and irrational numbers are provided.

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