Initial value of time function

Question

I need help find the the initial value for the following time function: $$F(s) = \frac{10(s+6)}{(s+2)(s+3)} $$

I've found the final value to be \$F(s)=0\$

Answer 1

Initial Value theorem States that the initial value is given by:

$$ \lim_{s \rightarrow \infty} sF(s) $$

so in our case, the initial value, \$f_0\$, is:

$$ f_0 = \lim_{s \rightarrow \infty}\frac{10s(s+6)}{(s+2)(s+3)} $$

$$ f_0 = \frac{10s^2}{s^2} = 10 $$