Let $f$ be a function defined as follows. $$f(x)= \cases{ x & if $x$ is irrational \\ 2a-x & if $x$ is rational}$$

Using the basic definitions of limits show that $\lim_{x\to x_0}f(x) $ does not exist, where $x_0>a$.

I tried to prove using contradiction. I assumed that limit exist and wrote for some $\delta>0$, $0<|x-x_0|<\delta \Rightarrow|f(x)-L|< \varepsilon$(any value can be put).

Then for some irrational $x_1$, $0<|x_1-x_0|<\delta \Rightarrow|f(x_1)-L|< \varepsilon$(any value can be put).

$|x_1-L|< \varepsilon$(any value can be put).

Then for some rational $x_2$, $0<|x_2-x_0|<\delta \Rightarrow|f(x_2)-L|< \varepsilon$(any value can be put).

$|2a-x_2-L|< \varepsilon$(any value can be put).

From these two, by triangle inequality, $|x_1+x_2-2a|< 2\varepsilon$.

Here, i cant find a suitable $\varepsilon$ to find the contradiction.

Can i proceed this way or i have to alter may method?

  • $\begingroup$ Isn't it easier to take two sequences converging to the same limit point, but one strictly in the rationals, and one striclty in the irrationals? If the limit would exists then the two limits should agree, but do they? Can you motivate that such sequences exist? $\endgroup$ – b00n heT Jun 18 '16 at 14:50
  • $\begingroup$ Here the question is to prove using the basic definion of limits. $\endgroup$ – user345749 Jun 18 '16 at 14:52

Suppose the limit is $k$ for some $x_0>a$. Then since $\epsilon=x_0-a>0$ we can find $\delta>0$ such that $|f(x)-k|<\epsilon$ for $|x-x_0|<\delta$.

We can find a rational $u\in(x_0,x_0+\delta)$ and an irrational $v\in(x_0,x_0+\delta)$. We have $f(u)=2a-u<2a-x_0$ and $f(v)=v>x_0$. So $f(v)>f(u)+2(x_0-a)=f(u)+2\epsilon$.

But since $|u-x_0|<\delta,|v-x_0|<\delta$ we also have $|f(u)-k|<\epsilon$ and $|f(v)-k|<\epsilon$, so $|f(u)-f(v)|<2\epsilon$. Contradiction.


Let's proceed using your approach. Then for every $\epsilon > 0$ there is $\delta > 0$ such that $$|f(x) - L| < \epsilon$$ for all $x$ with $0 < |x - x_{0}| < \delta$. Here $x_{0} > a$ and for every $\delta > 0$ we can choose rational $x_{1}$ and an irrational $x_{2}$ both greater than $x_{0}$ such that $0 < |x_{1} - x_{0}| < \delta$ and $0 < |x_{2} - x_{0}| < \delta$ and hence we have $$|f(x_{1}) - L| < \epsilon, |f(x_{2}) - L| < \epsilon$$ or $$|2a - x_{1} - L| < \epsilon, |x_{2} - L| < \epsilon$$ We can see that $$x_{2} - (2a - x_{1}) = |x_{2} - (2a - x_{1})| = |x_{2} - L + L - (2a - x_{1})|$$ and hence by triangle inequality $$x_{2} - (2a - x_{1}) < |x_{2} - L| + |L - (2a - x_{1})| < 2\epsilon\tag{1}$$ Now both $x_{1}, x_{2}$ are greater than $x_{0}$ and therefore both of them are greater than $a$ so that $2a - x_{1} < a$. Therefore $$x_{2} - (2a - x_{1}) > x_{2} - a = (x_{2} - x_{0}) + (x_{0} - a) > x_{0} - a\tag{2}$$ Clearly $(1)$ and $(2)$ contradict each other if $0 < \epsilon < (x_{0} - a)/2$. Thus our assumption is wrong and limit $L$ does not exist.

In the above proof I have chosen both $x_{1}, x_{2}$ to be greater than $x_{0}$, but this is not necessary (choosing them greater than $x_{0}$ only deals with limit $x \to x_{0}^{+}$). All that is necessary is to understand that we can choose $x_{1}, x_{2}$ so near to $x_{0}$ that there are both greater than $(a + x_{0})/2$ which is itself greater than $a$ and less than $x_{0}$. In fact we need any specific number $\xi$ which lies between $a$ and $x_{0}$ and we need to ensure that $x_{1}, x_{2}$ are greater than $\xi$. In that case we just have to choose $\epsilon < (\xi - a)$ and then we can obtain the contradiction needed because of the inequality $$x_{1} + x_{2} - 2a > 2(\xi - a)$$


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