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Calculate limit of piecewise function

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data
    a.) Let ##f,g:ℝ→ℝ## such that ##g(x)=sin x## and ##f(x)= \left\{
    \begin{array}{ll}
    x^2, x∈ℚ \\
    0 , x∈ℝ\setminusℚ \\
    \end{array}
    \right. ##. Calculate ##\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}##.

    b.) Why l'Hospital rule cannot be applied here?


    3. The attempt at a solution
    ##\lim_{x \rightarrow 0} sinx=0##


    ##\lim_{x \rightarrow 0} f(x)
    =\left\{\begin{array}{ll}
    0, x∈ℚ \\
    0 , x∈ℝ\setminusℚ \\
    \end{array}
    \right. ##


    How to begin? I suppose I cannot apply l'Hospital.
     
    Last edited: Apr 24, 2016
  2. jcsd
  3. Apr 24, 2016 #2

    andrewkirk

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    If you think about the values of ##f(x)## just on rational values of ##x##, it should become apparent what the limit must be, if there is a limit. Call that number ##l##. Then prove that ##l## actually is the limit by an explicit ##\epsilon##-##\delta## argument.
     
  4. Apr 24, 2016 #3
    The limit of ##\frac{f(x)}{g(x)}## is 0, right?
     
    Last edited: Apr 24, 2016
  5. Apr 24, 2016 #4

    Ray Vickson

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    l'Hospital's rule does not apply because ##f(x)## is not a differentiable function.
     
  6. Apr 24, 2016 #5
    Yes, ##f## is continuous and differentiable only at x=0 which is not sufficient.
     
    Last edited: Apr 24, 2016
  7. Apr 24, 2016 #6

    Ray Vickson

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    Right, because l'Hospital's rule talks about ##\lim_{x \to 0} f'(x)##, not just ##f'(0)## (at least as it is usually presented).
     
  8. Apr 24, 2016 #7

    andrewkirk

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    You are correct. So now you need to prove it, using an ##\epsilon##-##\delta## argument. That is, prove that, for any ##\epsilon>0## you can find a ##\delta>0## such that ##|x|<\delta\Rightarrow\frac{f(x)}{g(x)}<\epsilon##.
     
  9. Apr 25, 2016 #8
    If ##x∈ℝ\setminusℚ##, then ##\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x \rightarrow 0}\frac{0}{sinx}=\lim_{x \rightarrow 0}0=0## That's trivially true.

    If ##x∈ℚ##, then ##\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x \rightarrow 0}\frac{x^2}{sinx}=0## (an educated guess)

    Proof: Let ##ε>0## arbitrarily. Let ##δ=+\sqrt(ε|sinx|)##

    ##|\frac{x^2}{sinx}|=\frac{x^2}{|sinx|}<ε##, when ##|x|<δ=\sqrt(ε|sinx|)##.

    My proof doesn't look very elegant, does it?
     
    Last edited: Apr 25, 2016
  10. Apr 25, 2016 #8

    andrewkirk

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    ##\delta## cannot be a function of ##x##, as ##x## is not defined in that context. You either need a more careful ##\epsilon##-##\delta## argument, or you can use l'Hopital on ##\frac{x^2}{\sin x}##, use that to infer an ##\epsilon##-##\delta## relationship and apply that to the restricted domain of ##\mathbb{Q}##.
     
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