Real Analysis - Limits and Continuity of Piecewise Function 
Let $f(x) = \begin{cases}0, \text{if $x \in \mathbb{Q}$} \\ 1, \text{if $x \in \mathbb{R}-\mathbb{Q}$} \end{cases}$
Show that, for any $x_0 \in \mathbb{R}$, $\lim \limits_{x \to x_0}$ $f(x)$ does not exist.
Show that $f$ is nowhere continuous, i.e. for any $x_0 \in \mathbb{R}$, $f$ is not continuous at $x_0$.

I understand that the graph of $f$ is constantly jumping between $0$ and $1$ as we progress down the x-axis; this means that the LHS and RHS of the limit are not equal (because of the constant jumping), and therefore the limit cannot exist. Now this is obviously not a proof, and I know I need to use the epsilon-delta definition of a limit to prove this. How do I formulate the LHS limit $\ne$ RHS limit to show this? Is that even the right approach?
For the second part, it follows from the first, right? Since for any $x_0 \in \mathbb{R}$, the limit does not exist, which means the limit does not exist anywhere for $f$, which means $f$ cannot be continuous?
 A: 
  
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*How do I formulate the LHS limit $\neq$ RHS limit to show this? Is that even the right approach?
  

Neither LHS limit nor RHS limit does not exist. So your idea does not make sense.
Instead try to show that the limit does not exist using definition. It is enough to show that for any $L$, we can find $\epsilon>0$ such that for all $\delta>0$, the condition "$0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\epsilon$" does not hold.
Let us try. Fix a real number $L$ and set $\epsilon=\tfrac{1}{3}$. We need to show that for all $\delta>0$, the condition "$0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\epsilon$" does not hold. To this end, it suffices to find an $x$ such that $0<|x-x_0|<\delta$ but $|f(x)-L|\geq\epsilon=\tfrac{1}{3}$.
We already know that there are infinitely many rational numbers and infinitely many irrational numbers between $x_0$ and $x_0+\delta$. If $L$ is closer to $0$ than $1$, choose $x$ an irrational number between $x_0$ and $x_0+\delta$ so that $f(x)=1$. Otherwise choose $x$ a rational number between $x_0$ and $x_0+\delta$ so that $f(x)=0$.
In both case we have $|f(x)-L|\geq\tfrac{1}{3}$. Finally we found $x$ such that $0<|x-x_0|<\delta$ but $|f(x)-L|\geq\epsilon=\tfrac{1}{3}$. Therefore any real number $L$ cannot be a limit of $f(x)$ at $x_0$.


  
*Your argument is right.
  

