## Calculus (3rd Edition)

$$\frac{-1}{2}$$
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 $$\lim _{s \rightarrow 0} \frac{1-\sqrt{S^2+1} }{s^2}=\lim _{s \rightarrow 0} \frac{1-\sqrt{S^2+1} }{s^2} \frac{1+\sqrt{S^2+1}}{1+\sqrt{S^2+1}}\\ =\lim _{s \rightarrow 0} \frac{1-(S^2+1) }{s^2(1+\sqrt{S^2+1})} =\lim _{s \rightarrow 0} \frac{-1 }{ 1+\sqrt{S^2+1}}= \frac{-1}{1+1}=\frac{-1}{2}.$$