Show that the limit of fuction doesnot exist 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?
 A: 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 $$x_{1} + x_{2} - 2a > 2(\xi - a)$$
A: 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.
