Proof on Infinite Limits If the function $f$ is defined by 
$f(x)=0$ if $x$ is rational
$f(x)=1$ if $x$ is irrational 
Prove that $\lim\limits_{x\to 0}f(x)$ does not exist. 
Can someone help me answer this question step by step so I would know how to prove a question like this next time? Thank you.
 A: Let $\{a_n\}$ be a sequence of rationals converging to $0$ and $\{b_n\}$ be a sequence of irrationals converging to $0$.
Then look at the sequence of functions for each sequence.
A: An equivalent approach to Yunus's is to go to the definition. Suppose the limit exists and is $L$. Then for any $\epsilon > 0$, there would exist a $\delta > 0$ such that $0 < |x| < \delta \implies |f(x) - L| < \epsilon$. But no matter how small $\delta$ is ,there are both rationals and irrationals that are less than it. So some values of $f(x)$ will be $0$ and others will be $1$. If $\epsilon < 0.5$, then there is no value of $L$ that will work for every $x$ less than $\delta$.
A: For any $x=a,$
$\bf{L.H.L(Left\; hand \; limit) = } \lim_{x\rightarrow a^{-1}f(x)} = \lim_{h\rightarrow 0}f(a-h) = 0$ or $1$
(as $\lim_{h\rightarrow 0}f(a-h)$ can be rational or irrational.) 
Similarly $\bf{R.H.L(Right\; hand \; limit) = } \lim_{x\rightarrow a^{+1}f(x)} = \lim_{h\rightarrow 0}f(a+h) = 0$ or $1$
Hence $f(x)$ oscillates between $0$ and $1$ as for all values of $a$
So $\bf{L.H.L(Left\; hand \; limit)}$ and $\bf{R.H.L(Right\; hand \; limit)  }$ does not exists
