So I have the following:
$G(s)=\dfrac{1}{(s+7)(s+12)}$, $ȳ(s)=\dfrac{x̄(s)}{(s+7)(s+12)}$
Since $x(t)=(1+1/t)^t$ which is $e$ as t tends towards infinity, can I simply say that $L(x(t))=x̄(s)=e/s$
The question was to calculate the limit of the output $y(t)$ as t tends towards infinity. Using the converging input converging output theorem, I have
$\lim_{t\rightarrow\infty}x(t)=x_\infty=e$
$\lim_{t\rightarrow\infty}y(t)=G(0)x_\infty=e/84$
Is this correct?
You can use the final value theorem which states that
\begin{equation} \lim_{t\rightarrow\infty}f(t)=\lim_{s\rightarrow0} s F(s) \end{equation}
where $F(s)$ is the Laplace transfort of $f(t)$. So \begin{equation} \lim_{s\rightarrow0} s \frac{e}{s} \dfrac{1}{(s+7)(s+12)} =\frac{e}{84} \end{equation}