Is it possible to find $\limsup\limits_{n\to\infty} \frac{2×3^n-3×2^n}{2^{\alpha(n)}-3^n}$? I need help to solve this problem:

$$\lim_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n}{2^{\alpha(n)}-3^n}$$ or
$$\limsup_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n } {2^{\alpha(n)}-3^n}$$

where,
$\alpha(n)-$ is the smallest positive integer, such that $2^{\alpha(n)}-3^n>0$.
I tried $\alpha (n)= \lfloor n\log_23+1 \rfloor$. Then, we have
$$\lim\limits_{n\to\infty}\frac{2\cdot 3^n-3\cdot 2^n}{2^{\alpha(n)}-3^n}= \lim\limits_{n\to\infty} \frac{2}{\frac{2^{\lfloor n\log_23+1 \rfloor}}{3^n}-1}$$
I'm stuck here. I can not continue..Can MSE help me? Are my steps wrong?
Can I take $\alpha (n)= \lfloor n\log_23+1 \rfloor$ ?
Thank you.
 A: We have
$$\alpha(n) = \lceil n \log_2 3 \rceil = n \log_2 3 + 1 - \{n \log_2 3\}, \\
\frac {2 \cdot 3^n - 3 \cdot 2^n} {2^{\alpha(n)} - 3^n} =
\underbrace {3^{-n} (3 \cdot 2^n - 2 \cdot 3^n)}_{\to -2} \cdot
 \frac 1 {1 - 2^{1 - \{n \log_2 3 \}}} \,.$$
Since $\{n \beta\}$ is dense in $[0, 1]$ for irrational $\beta$, we can choose a subsequence that doesn't converge and can choose a subsequence that converges to $1$.
A: I don't think the limit exists. Write $3^n=2^{n\log_2(3)}$, then the latter simplifies to
$$ \frac{2}{2^{\lfloor n\log_2(3) + 1 \rfloor - n\log_2(3)}-1} =  \frac{2}{2^{\lfloor n\log_2(3) + 1 \rfloor - n\log_2(3)}-1} $$
I believe there exists an increasing sequence $(a_n)_{n=1}^{\infty}\in\mathbb{N}^{\mathbb{N}}$ such that
$\lfloor a_n\log_2(3) + 1 \rfloor - a_n\log_2(3)\to 0$. The limit under this subsequence will thus converge to infinity. Moreover, we can also pick the $(a_n)$ such that $\lfloor a_n\log_2(3) + 1 \rfloor - a_n\log_2(3)\to1$, which means the subsequence converges to $1$. Thus, the limit can't exist.
I am $99\%$ sure that this is a valid argument. However, to be precise, you have to define the $a_n$ in both cases and work out their respective limits.
