Q:

Show that f is discontinuous in 1.

Use $\epsilon$-$\delta$ definition to show that $f(x)=\frac{x-1}{|x-1|}$ is discontinues in 1.
I'm not sure how I'm supposed to solve this. Can someone give me a good explanation?

A:

$$\lim_{\epsilon_+\to 0}f(x+ \epsilon)|_{x=1}=\frac{x+\epsilon-1}{|x+\epsilon-1|}=\frac{\epsilon}{|\epsilon|}=+ 1$$
$$\lim_{\epsilon_-\to 0}f(x- \epsilon)|_{x=1}=\frac{x-\epsilon-1}{|x-\epsilon-1|}=\frac{-\epsilon}{|\epsilon|}= -1$$
$f(x+ \epsilon) \neq f(x- \epsilon)$ , so it is discontinues at  $x=1$.

