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My problem is :Find the points at which the the mentioned function is continuous $$f(x) = \begin{cases} x & \text{if $x$ is a Rational Number} \\ -x & \text{if $x$ is not a Rational Number} \end{cases}$$ I was asked to learn that this function is continuous at $x = 0$ and the LHL and RHL were equated as follows

LHL $$ \lim_{h \to 0} f(0 - h) = \lim_{h \to 0}-(0 - h) = 0$$ ,

RHL $$ \lim_{h \to 0} f(0 + h) = \lim_{h \to 0}-(0 + h) = 0$$ and

$$f(0) = 0$$ Now since $$ LHL = RHL = f(0)$$ Therefore the function is continuous.

My question is why are we taking a point just before Zero to be Irrational. In my opinion it could be Rational as well as Irrational making the function oscillatory and hence making it discontinuous.Please help.

I possible please state your educational qualifications(It will help me when I discuss the solution with my teacher).

Normal Boy
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    You are right, the proof here is non-sense. There are both rational and irrational points arbitrarily close $0$, so it will be oscillating as it approaches zero. But the function is continuous at zero, because the amplitude of the oscillations go to zero. – EuYu Jun 07 '16 at 06:07
  • Yes. In both the LHL and the RHL one has to consider rationals and irrationals. One way is to choose any sequence converging to zero and then consider the particular cases. However, the easiest way here is probably the epsilon-delta formulation of continuity. – Friedrich Philipp Jun 07 '16 at 06:12
  • sorry but i dont know epsilon-delta formulation of continuity. – Normal Boy Jun 07 '16 at 06:29

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The given argument is no proof: it's only a complicated way to show the rather obvious fact that $x\mapsto -x$ is continuous at $0$.

What you want is showing that $\lim_{x\to0}f(x)=0$ and this follows easily from the fact that $$ -|x|\le f(x)\le |x| $$ for all $x$.

Apply the squeeze theorem.

No squeeze theorem? Then let's go with the definitions.

Let $\varepsilon>0$; if $0<|x-0|<\varepsilon$, then $|f(x)-0|=|f(x)|=|x|<\varepsilon$. So taking $\delta=\varepsilon$ ends the argument.

egreg
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  • I am sorry but I really dont know the squeeze theorem. Thanks anyway! – Normal Boy Jun 07 '16 at 08:39
  • @NormalBoy I added the $\varepsilon$-$\delta$ proof – egreg Jun 07 '16 at 09:01
  • I am really sorry but high school maths does not include the epsilon delta proof. I hope you can give an even simpler and better answer. – Normal Boy Jun 07 '16 at 15:20
  • @NormalBoy One can talk a bit more loosely about continuity, but without formal definitions it is just small talk that leads to nothing. – egreg Jun 07 '16 at 15:25
  • I understand your point, but right now I am reaching nowhere. I still feel the function is discontinuous in the neighborhood of 0.(Rational or irRational confusion). – Normal Boy Jun 10 '16 at 16:48
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Limit value should be same irrespective of the way x approaches a (=0 in this case). Here it is totally incorrect to apply RHL and LHL. Correct way would be as follows: To ensure continuity at 0 we should have lim x tends to 0, such that x belongs to rational f(x)= lim x tends to 0, such that x belongs to irrational f(x)= f(0). Note that we can do this since we can always find a rational as well as irrational number AS CLOSE to any number (0 here) AS WE PLEASE. So, lim x tends to 0, such that x belongs to rational x= lim x tends to 0, such that x belongs to irrational -x= f(0) Or 0= -0= 0 which is clearly true. Hence proved.

aditya gupta
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