Use initial value theorem to find f(0)

Question : Use initial value theorem to find f(0) for F(s)=(3+4s^2+5s)(s(s^2+4s+2))

Solution :

We shall Use initial value theorem to find f(0) for F(s) = F(s)=(3+4s^2+5s)(s(s^2+4s+2))

Given function \[F\left(s\right)=\frac{3+4s^2+5s}{s\left(s^2+4s+2\right)}\] We need to write the function sF(s) as \[sF\left(s\right)=\frac{3+4s^2+5s}{s^2+4s+2}\] Now by initial value theorem \[f\left(0\right)=limit_{s\rightarrow\infty\ }sF\left(s\right) \]Substituting sF(s) \[f\left(0\right)=limit_{s\rightarrow\infty\ }\frac{3+4s^2+5s}{s^2+4s+2}\] We get \[f\left(0\right)=limit_{s\rightarrow\infty\ }\frac{\frac{3}{s^2}+4+\frac{5}{s}}{1+\frac{4}{s}+\frac{2}{s^2}}\]\[=\frac{4}{1}=4\] Thus Using initial value theorem the value of f(0) is 4