Let's say this is part of an epsilon N proof : $|3^{n+1} - 5\cdot 2^{n-5}|<\epsilon$.
If we have $n$-th powers in equation how do we generally go about ending the proof?
This inequality is part of the proof for :
$$e_n = \frac{4\cdot3^{n+1} + 2\cdot4^n}{5\cdot2^n+4^{n+2}}$$
$$\begin{align} \lim_{n \to \infty} \frac{4\cdot3^{n+1} + 2\cdot4^n}{5\cdot2^n+4^{n+2}} &= \lim_{n \to \infty} \frac{\frac{3^{n+1}}{4^{n+1}} + \frac{2}{4^2}}{5\cdot\frac{2^n}{4^{n+2}}+1}\\ &=\lim_{n \to \infty}e_n = \frac{1}{8} \end{align}$$
If we set $$\left|\frac{4\cdot3^{n+1} + 2\cdot4^n}{5\cdot2^n+4^{n+2}} - \frac{1}{8} \right|<\epsilon$$ After some transformations we arrive with $|3^{n+1} - 5\cdot2^{n-5}|<\epsilon$.
You did not divide by $4^n+$ other terms. That makes it easier.