<s>[INST] Limit of $\left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- 1/ {n^4}} {4+ n^7/ {3^n}}$ as $n$ tends to infinity I already took some steps to simplifying the original question and im stuck at this point:
$$ \lim_{n \to \infty} \left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- \frac {1} {n^4}} {4+ \frac {n^7} {3^n}} $$ 
I know that $$\lim_{n \to \infty}\left(\frac {2}{3}\right)^n=0$$
since $$0<\frac{2}{3}<1$$  
and that $$\lim_{n \to \infty}\frac {1} {n^4}=0$$
and that I have to use d'Alembert's criterion to solve what $$\frac {n^7} {3^n}$$ is
but what I get is $$\frac 13*\frac {(n+7)^7}{n^7}$$
I suppose the next step is just $$\frac 13 <1$$
"therefore the limit is zero" but my question is how do I get rid of the
$$\frac {(n+7)^7}{n^7}$$
The last step is just:
$$0*\frac {1-0}{4+0}=0$$ 
 [/INST] HINT:
We have $$\frac{2^n(n^4-1)}{4\cdot3^n+n^7}<\frac{2^n}{4\cdot3^n}<\frac14\left(\frac23\right)^n$$
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