How you'd show that $f$ is not continuous? How do you show that 
$$f(x)=\begin{cases}2,&x < c\\1,&x\geq c\end{cases}$$ is not continuous at $c$ by using $\epsilon$-$\delta$ - formalism like here?
 A: Suppose $f$ were continuous.  Let $\epsilon=0.1$, and $\delta$ be chosen so that $|f(x)-f(c)|<\epsilon$ for all $|x-c|<\delta$. But $|(c-\delta/2)-c|<\delta$ and $|f(c-\delta/2)-f(c)|=|2-1|=1>\epsilon$.  Contradiction.
A: Hint: You must show that there exists an $\epsilon>0$ so that no matter what $\delta>0$ you choose, $0<\lvert x-c\rvert<\delta$ is not enough to say that $\lvert f(x)-f(c)\rvert<\epsilon$.  
That is, there exists an $\epsilon>0$ so that for any $\delta>0$, you can find a point $x$ which is within $\delta$ of $c$ but satisfies $\lvert f(x)-f(c)\rvert\geq\epsilon$.
You can find points as close to $c$ as you want which have $f(x)=2$, so that $\lvert f(x)-f(c)\rvert=1$.  How should you choose $\epsilon$?  
A: You need to show that there exists some $\epsilon>0$ such that for any $\delta>0$, you can find some $x$ with $|x-c|< \delta$ so that $|f(x)-f(c)| \ge \epsilon$.
In this particular case, this means finding some $\epsilon>0$ such that for any $\delta>0$, you can find some $x$ with $|x-c|< \delta$ so that $|f(x)-1| \ge \epsilon$.
If you draw a picture it may help find appropriate $\epsilon$ and $x$.
A: Reword the definition:
$f$ is not continuous at $x_0$ if there exists an $\epsilon > 0$ such that for every $\delta > 0$, there exists an $x$ such that $|x-x_0| < \delta$ but $|f(x) - f(x_0)| > \epsilon$.
Now, let $x_0 = c$, take $\epsilon = 1/2$ (other choices will work), and then show that for every $\delta > 0$, there exists an $x$...
