Answer
$$\frac{-1}{2}$$
Work Step by Step
By substituting, we get the intermediate form $\frac{0}{0}$. So we multiply the function by $ \frac{1+\sqrt{S^2+1}}{1+\sqrt{S^2+1}}$; then we have
$$
\lim _{s \rightarrow 0} \frac{1-\sqrt{S^2+1} }{s^2}=\lim _{s \rightarrow 0} \frac{1-\sqrt{S^2+1} }{s^2} \frac{1+\sqrt{S^2+1}}{1+\sqrt{S^2+1}}\\
=\lim _{s \rightarrow 0} \frac{1-(S^2+1) }{s^2(1+\sqrt{S^2+1})} =\lim _{s \rightarrow 0} \frac{-1 }{ 1+\sqrt{S^2+1}}=
\frac{-1}{1+1}=\frac{-1}{2}.
$$
