Answer
a. (i) 1 (ii) -1 (iii) DNE (iv) -1 (v) 1 (vi) DNE
b. $lim_{x\to a}sgn\left(sinx\right)=DNE$ if $a=n\pi$
c. See graph
Work Step by Step
a. $0^{+}$: means from right to 0 (small positive number)
$0^{-}$: means from left to 0 (negative positive number)
(i) $\lim _{x\to 0^+}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to \:0^+}sinx\right)=sgn\left(0^+\right)=1$
(ii) $\lim _{x\to 0^-}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to \:0^-}sinx\right)=sgn\left(0^-\right)=-1$
(iii) we have to look at (i) and (ii)
$\lim _{x\to 0^+}sgn\left(sinx\right)\ne\lim _{x\to 0^-}sgn\left(sinx\right)$
$\lim _{x\to 0}sgn\left(sinx\right)=DNE$
(iv) $\lim _{x\to \:\pi \:^+\:}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to\pi \:^+\:}sinx\right)=sgn\left(0^-\right)=-1$
(v) $\lim _{x\to \:\pi \:^-}sgn\left(sinx\right)=sgn\left(\lim \:_{x\to \pi \:^-}sinx\right)=sgn\left(0^+\right)=1$
(vi) we have to look (iv) and (v)
$lim_{x\to \:\pi ^+\:}sgn\left(sinx\right)\ne lim_{x\to\pi \:^-}\:sgn\left(sinx\right)$
$lim_{x\to \:\pi}sgn\left(sinx\right)=DNE$
b. $\lim _{x\to a}sgn\left(sin(x)\right)$ does not exist when $a=0$
then $\lim _{x\to a}sgn\left(sin(x)\right)$ not exist when $sina=0$
$sin a=0$ when $a=n\pi$
$lim_{x\to a}sgn\left(sinx\right)=DNE$ if $a=n\pi$
c.See graph
