## Calculus: Early Transcendentals 8th Edition

1. $x\rightarrow 0^{+}$ from right of 0, sgn x=1 $\lim _{x\to 0^+}\left(sgn\:x\right)=1$ 2. $x\rightarrow 0^{-}$ from left of 0, sgn x=-1 $\lim _{x\to 0^+}\left(sgn\:x\right)=-1$ 3. As we know from (1) and (2) $\lim _{x\to 0^+}\left(sgn\:x\right)=1 \ne \lim _{x\to 0^+}\left(sgn\:x\right)=-1$ Since the two sides do not equal, limit does not exist. $\lim _{x\to 0}\left(sgn\:x\right)=D.N.E.$ 4. $\lim _{x\to 0^+}\left(\left|sgnx\right|\right)=1$ $\lim _{x\to 0^-}\left(\left|sgnx\right|\right)=1$ Since two sides are equal then $\lim _{x\to 0}\left(\left|sgnx\right|\right)=1$