Finding the limit of $\frac{3^{n+1}}{4^{n-1}}$ as $n\rightarrow\infty$ How would one find the limit of the following.
as $n\rightarrow\infty$
$$\frac{3^{n+1}}{4^{n-1}}$$
I did the following
$$\frac{3\cdot 3^m}{1/4 \cdot 4^m}$$
$$\frac{3}{\frac{1}{4}}\ \frac{3^m}{4^m}$$
The limit if zero.
Another one I did is
$$\frac{2^{3n-1}}{7^{n+2}}$$
I did the following
$$\frac{\frac{1}{2}}{49}\ \frac{8^n}{7^n}$$
which means n approach infinity
But I am not sure if I did this correctly the numbers I got on the let side.

$\lim_{n\to\infty}\frac{3^{n+1}}{4^{n-1}}=\lim_{n\to\infty}\frac{3^{n}\cdot 3}{4^{n}\cdot 4^{-1}}=\lim_{n\to\infty}\frac{3^{n}\cdot 3\cdot 4}{4^{n}}=12\lim_{n\to\infty}(\frac{3}{4})^{n}$=$12\cdot 0$=$0$, because $\lim_{n\to\infty}(\frac{3}{4})^{n}=0$.
Your answer is correct.
and the another
$\lim_{n\to\infty}\frac{2^{3n-1}}{7^{n+2}}=\lim_{n\to\infty}\frac{2^{3n}\cdot 2^{-1}}{7^{n}\cdot 7^{2}}=\lim_{n\to\infty}\frac{2^{3n}}{7^{n}\cdot 7^2\cdot 2}=\frac{1}{98}\lim_{n\to\infty}(\frac{8}{7})^{n}$=$\infty$, because $\lim_{n\to\infty}(\frac{8}{7})^{n}=\infty$.
