$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$ How do I solve this example? I tried to point out the fastest growing term $2 ^ n$ and $3 ^ n$, but that doesn't seem to lead to the result. I know the limit is $0$ that's obvious, but I don't know how to work on it.
$$\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$$
Or can I make a power estimate of a theorem on a tightened sequence, for example, that $2 ^ n$ / $13 ^ n$ (smaller) goes to zero and the other side  $2 ^ n$ / $3 ^ n$ (bigger) also goes to zero?
$\lim_{n\to\infty} 2 ^ n$ / $13 ^ n$< $\lim_{n\to\infty} \frac{2^n(n^4-1)}{4\cdot 3^n + n^7}$ < $\lim_{n\to\infty} 2 ^ n$ / $3 ^ n$

It follows from the ratio test.

If $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = a < 1 \ $, then $\lim a_n = 0$.

$$ \dfrac{\frac{2^{n+1} \left((n+1)^4-1\right)}{4\cdot3^{n+1} + (n+1)^7}}{\frac{2^n(n^4-1)}{4\cdot3^n + n^7}} = \dfrac{2^{n+1} \left((n+1)^4-1\right)(4\cdot3^n + n^7)}{2^n(n^4-1)\left(4\cdot3^{n+1} + (n+1)^7\right)} \\ =\frac{2}{3} \left(\frac{n+1}{n}\right)^4 \left(\dfrac{1-\frac{1}{(n+1)^4}}{1-\frac{1}{n^4}}\right) \left(\dfrac{1+\frac{n^7}{3^n}}{1+\frac{(n+1)^7}{3^{n+1}}} \right) \longrightarrow \frac{2}{3} < 1 $$
