Answer
$$3$$
Work Step by Step
Given $$ a_{n}=\frac{1}{2} 3^{n}-\frac{1}{3} 2^{n} $$
Since
$$ a_{n+1}=\frac{1}{2} 3^{n+1}-\frac{1}{3} 2^{n+1}$$
Then
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}&=\lim _{n \rightarrow \infty} \frac{\frac{1}{2} 3^{n+1}-\frac{1}{3} 2^{n+1}}{\frac{1}{2} 3^{n}-\frac{1}{3} 2^{n}}\\
&=\lim _{n \rightarrow \infty} \frac{3^{n+2}-2^{n+2}}{3^{n+1}-2^{n+1}}\\
&=\lim _{n \rightarrow \infty} \frac{3-2\left(\frac{2}{3}\right)^{n+1}}{1-\left(\frac{2}{3}\right)^{n+1}}\\
&=\frac{3-0}{1-0}=3
\end{align*}
