Answer
Yes, the infinite geometric series has a limit, and the limit is $48$.
Work Step by Step
The provided series is $12++9+\frac{27}{4}+\cdots $.
Here, ${{a}_{1}}=12$, $n=\infty $ and
$\begin{align}
& r=\frac{9}{12} \\
& =\frac{3}{4}
\end{align}$
Substituting ${{a}_{1}}=12$, $n=\infty $ and $r=\frac{3}{4}$ in ${{S}_{n}}=\frac{{{a}_{1}}}{1-r}$ , for any $r\ne 1$:
$\begin{align}
& {{S}_{\infty }}=\frac{12}{1-\frac{3}{4}} \\
& =\frac{12}{\frac{1}{4}} \\
& =\frac{12\cdot 4}{1} \\
& =48
\end{align}$
Thus, the infinite geometric series $12++9+\frac{27}{4}+\cdots $has a limit of 48.
