Answer
$\lim\limits_{s\to-2}f(s)=2.$
Work Step by Step
$\lim\limits_{s\to-2^-}f(s)=\lim\limits_{s\to-2^-}(-s^2-4s-2)=-(-2^-)^2-4(-2^-)-2=2.$
$\lim\limits_{s\to-2^+}f(s)=\lim\limits_{s\to-2^+}(s^2+4s+6)=(-2^+)^2+4(-2^+)+6=2.$
Since $\lim\limits_{s\to-2^-}f(s)=\lim\limits_{s\to-2^+}f(s)\to\lim\limits_{s\to-2}f(s)$ exists and is equal to $2.$
