$\displaystyle \lim_{t\rightarrow 0}\ f(t)= \lim_{t \to \infty}[20-10t-e^{-25t}] = 20-1 =19$

$f\left(t\right)=20-10t-e^{-25t} $

Taking Laplace transform

$f\left(s\right)=\dfrac{20}{s}-\dfrac{10}{s^2}-\dfrac{1}{s+25}\\[3ex] s\left(f\right)=20-\dfrac{10}{s}-\dfrac{s}{s+25} $

By initial value theorem

$\displaystyle \lim_{s\rightarrow \infty}\ f(t)= \lim_{s \to \infty}\Bigg[20-\dfrac{10}{s}-\dfrac{s}{s+25}\Bigg] = 20-1 =19$