How to evaluate limit of a sequence $ \lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}} $

Question

I need a help with evaluating a limit of a sequence $$ \lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}}. $$ The problem is that $\lim_{n \to \infty} \left( -2 \right)^n$ does not exist. What can we even tell from this when there's a part oscilating between $+\infty$ and $- \infty$. Wolfram says it should equal to $ - \infty$, but how to get there? The only thing I know might help is to factor out the fastest growing terms.

$$ \lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}} = \frac{1}{3}\lim_{n \to \infty} \frac {2 \cdot 3^{2n} - \left( -2 \right)^n}{2 \cdot 3^n - 6 \cdot 2^{2n}} = \frac{1}{3} \lim_{n \to \infty} \left( \frac {3}{2} \right)^{2n} \cdot \frac {2 - \left( - \frac {2} {9} \right)^n}{2 \cdot \left( \frac{3}{4} \right)^n - 6} $$ Now we can see that $\left( \frac {3} {4} \right)^n $ goes to zero. What about $\left(- \frac{2}{9} \right)^n $? I guess that despite the fact that the values oscilate between $+$ and $-$, the overall fraction has to go to zero, hence giving $$\left| \frac{1}{3} \cdot \infty \cdot \frac{2}{-6} \right| = \left| - \frac{\infty}{9} \right| = -\infty$$ Does that make sense?

Answer 1

Divide by $3^{2n}$, numerator becomes $\frac{2}{3} - (-1)^n (\frac{2}{3})^n$, so the second term converges to 0. In he denominator you also get term that converges to $0$ - another term that converges to $0$, so in total the expression has limit of the type $\frac{1-0}{0-0}$