Answer
(a)$$\lim_{x \to 0^-}sgn(x)=-1$$
(b)$$\lim_{x \to 0^+}sgn(x)=1$$
(c) $\lim_{x \to 0}sgn(x)$ does not exist.
Work Step by Step
As we see the graph of $sgn(x)$, we can easily find that:$$\lim_{x \to 0^-}sgn(x)=-1 \quad \& \quad \lim_{x \to 0^+}sgn(x)=1 .$$ The left- and right-hand limits at $x=0$ are different. Thus, $\lim_{ x \to 0}sgn(x)$ does not exist, according to Theorem 1.10.
