# Thread: question regarding continuity of sign function

1. ## question regarding continuity of sign function

Hi! I need to show that the sign function:
$f(x)=\left\{\begin{array}{lr}1:&x>0\\0:&x=0\\-1:&x<0\end{array}\right\}$
is discontinuous.
Is it true that $\lim_{x\to 0^-}f(x)=1$ and $\lim_{x\to 0^+}f(x)=-1$ so therefore $\lim_{x\to 0}f(x)$ does not exist? Im confused because doesnt the limit of f(x) at x=0 exist and is equal to 0?
Does that somehow mean that f(x) is continuous at 0 but discontinous at 0?

2. Originally Posted by dannyboycurtis
Hi! I need to show that the sign function:
$f(x)=\left\{\begin{array}{lr}1:&x>0\\0:&x=0\\-1:&x<0\end{array}\right\}$
is discontinuous.
Is it true that $\lim_{x\to 0^-}f(x)=1$ and $\lim_{x\to 0^+}f(x)=-1$ so therefore $\lim_{x\to 0}f(x)$ does not exist? Im confused because doesnt the limit of f(x) at x=0 exist and is equal to 0?
Does that somehow mean that f(x) is continuous at 0 but discontinous at 0?

Yes, no and no: the limit of a function at some point exists iff both one-sided (or lateral) limits exist at that point and are equal, so yes.
The limit of f(x) at x = 0 doesn't exist since $f(x)\xrightarrow [x \to 0^{-}] {} -1\,,\,\,and\,\,f(x)\xrightarrow [x \to 0^{+}] {} 1$ , and so no.
Finally, as $\lim_{x\rightarrow 0}f(x)\,$ doesn't exist f(x) cannot be continuous at 0 and thus no.

Tonio