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Homework Help: [Calculus]Hard continutity problem.

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Define [itex]G[/itex] as follows:

    [itex]G(x) = \left\{
    \begin{array}{c l}
    x, & \mbox{if } x \mbox{ is irrational} \\
    \sqrt{\frac{1+p^2}{1+q^2}}, & \mbox{if } x = \frac{p}{q} \mbox{where } gcd(p,q) = 1

    Show that [itex]G[/itex] is discontinuous at each negative number and also at each nonnegative rational number, but is continuous at each positive irrational number.

    2. Relevant equations
    [itex]\lim_{x \rightarrow a} f(x) = f(a)[/itex]
    3. The attempt at a solution
    I found this question in an old calculus book at the end of the limits and continuity section.
    I tried the approach using [itex]\lim_{x \rightarrow a} G(x) = G(a)[/itex] and trying the different cases for [itex]a[/itex] and it makes sense intuitively but I'm thinking and epsilon-delta approach is what is needed here. Any help?
    Last edited: Apr 6, 2012
  2. jcsd
  3. Apr 6, 2012 #2


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    Show what your reasoning is so far.
  4. Apr 6, 2012 #3
    Sorry I didn't put it before. It takes a while to post from my phone.

    I first test the case where x goes to an irrational number. Say b.
    G(b) = b, so now we must look at the limit as x--->b of G(x).
    My reasoning is that there is rationals around G(b) that approach this value as x gets closer and closer to b.

    So for nonnegative irrationals continuity holds.

    For the cases of positive or negative rationals we let wlog some number r = s/t where s and t are coprime.

    Then G(r) = √(1+s^2)/(1+t^2). Now the limit as x--->r of G(x) wouldn't exist as x has to go through irrational values as it gets closer to r from either positive or negative side. So it wouldn't be continuous as
    lim as x--->r of G(x) is not equal to G(r).
    Last edited: Apr 6, 2012
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