$f(x) = \begin{cases} -1, & \text{if $x<0$} \\[2ex] 0, & \text{if $x = 0$} \\[2ex] 1, & \text{if $x>0$} \end{cases}$
How do I prove that the limit $\lim_{x\to0}f(x)$ doesn't exist using epsilon-delta definition?
I don't know how to proceed from $|f(x) - L| <\epsilon$, since I don't know how to define what $f(x)$ is when $x$ is getting close to $0$
Show $\lim_{x \to o^+} f(x) = 1, \lim_{x \to 0^-} f(x)=-1$.