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:
$$0*\frac {1-0}{4+0}=0$$ 
 A: I think you can get this by doing a bit of algebra. Since $1-\frac{1}{n^4}<4+\frac{n^7}{3^n}$ for all $n \in \Bbb{N}$ then $$ \left(\frac{2}{3}\right)^n \cdot n^4 \cdot \frac{1- \frac{1}{n^4}} {4+ \frac{n^7}{3^n}}< \left(\frac{2}{3}\right)^n \cdot n^4$$ Now let's use the fact that exponents grow faster than powers. We know that there exists some $M \in \Bbb{N}$ that $$\left(\frac{4}{3} \right)^m>m^4 \quad \text{for all} \quad m>M$$ which means $$\left(\frac{2}{3}\right)^m \cdot m^4< \left(\frac{2}{3}\right)^m\left(\frac{4}{3} \right)^m \\ = \left(\frac{8}{9} \right)^m$$ and we know that $\lim_{m \to \infty} \left(\frac{8}{9} \right)^m=0$. Thus, for all $m>M$  $$\lim_{m \to \infty} \left(\frac{2}{3}\right)^m \cdot m^4 \cdot \frac{1- \frac{1}{m^4}}{4+ \frac{m^7}{3^m}} < \lim_{m \to \infty}\left(\frac{8}{9} \right)^m$$ which means by the comparison test that $$\lim_{m \to \infty} \left(\frac{2}{3}\right)^m \cdot m^4 \cdot \frac{1- \frac{1}{m^4}}{4+ \frac{m^7}{3^m}}=0$$
A: Hint: for any $m>0$, $d\in (0,1)$, it holds $n^m/d^n\to 0$...
A: HINT:
We have $$\frac{2^n(n^4-1)}{4\cdot3^n+n^7}<\frac{2^n}{4\cdot3^n}<\frac14\left(\frac23\right)^n$$
