Solution:

Initial value:

$x(0)=\underset{s \rightarrow \infty}{\mathrm{Lt}} S X(S)\\ $

$=\operatorname{Lt}_{s \rightarrow \infty} s \frac{1}{s(s+2)}=\frac{1}{\infty}=0\\ $

Final value,

$x(\infty)=\underset{s \rightarrow 0}{\operatorname{Lt}} S X(S)\\ $

$=\operatorname{Lt}_{s \rightarrow 0} s \frac{1}{s(s+2)}=\frac{1}{2}$